Atomic Structure Exercises by Resonance

ATOMIC STRUCTURE # 37

Type (I) : Very Short Answer Type Questions : [01 Mark Each]

1. Calculate the number of electrons, protons and neutrons in the following species :

(i) P3115 (ii) 331

15P

2. What is the rest mass of a photon ?

3. What is the energy of the electron in He+ in ground state ?

4. Calculate the total number of electrons in 1 mole of ammonia.

5. What is the difference between a photon and a quantum ?

6. The number of electrons, protons and neutrons in a monoatomic species are equal to 36, 35 and 45 respectively.Assign the proper symbol.

7. Give an isobar, an isotone and an isotope of C146 .

8. The azimuthal quantum number of an orbital is 3. What are the possible values of m ?

9. How many electrons in a fully filled f-subshell have m = 0 ?

10. Why 3d sub-shell has higher energy than 4s sub-shell in a muti-electron atom ?

Type (II) : Short Answer Type Questions : [02 Marks Each]

11. Arrange the following type of radiations in increasing order of frequency :(a) radiation from microwave oven (b) amber light from traffic signal, (c) radiation from FM radio, (d) cosmicrays from outer space and (e) X-rays

12. Calculate the energy required for the process :He+(g) He2+(g) + e–

The ionization energy for the H atom in the ground state is 2.18 × 10–18 J/atom.

13. Calculate the wavelength of a particle of mass m = 6.6 × 10–27 kg moving with kinetic energy 7.425 × 10–13 J.(h = 6.6 × 10–34 kg m2s–1).

14. If an electron is to be located within 10 pm what will be uncertainty in its velocity ?

15. (a) How many sub-shells are there in N shell ?(b) How many orbitals are there in d sub-shell ?

16. Consider the following radioactive change NeNa 2210

2211

What is the type of particle captured or emitted ?

17. How many orbitals are possible in(i) 4th energy level(ii) 5f-sub-shell ?

18. How many spherical nodal surfaces are there in (i) 3s orbital, (ii) 3p orbital ?

19. What is the number of d electrons in Cr3+ ions ?

20. Calculate the mass of a photon having wavelength 1 nm.

21. How many quantum numbers are needed to designate an orbital ? Name them.


22. Calculate the number of and -particles emitted in the change :

PbU 20782

23592

Type (III) : Long Answer Type Questions: [03 Mark Each]

23. What is the shape of 2s-orbital ? Give two points of difference between 1s and 2s orbitals.

24. Write electronic configuration of Fe2+ and Fe3+ ions. Which of these has more number of unpaired electrons?Atomic no. of Fe is 26.

25. Derive the relationship between the wavelength of the de-Broglie wave and the kinetic energy of the particle.

26. State and explain Heisenberg's uncertainty principle.

27. Draw the shapes of various p and d orbitals.

28. Write balanced nuclear equations for each of the following :(a) -emission from curium-242(b) -emission from magnesium-28(c) positron emission from xenon-118

29. Among the following pairs of orbitals which orbitals will experience the larger effective nuclear charge ?(i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p

30. State Hund's rule of maximum multiplicity. How is it used in electronic distribution in nitrogen atom (Z = 7) ?

31. Describe the experiment which led to the discovery of the proton.

32. What are the shortcomings of Rutherford's model of atom ?

33. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024

per second, calculate the power of this laser.

34. Define the followings :(i) Nuclear fission (ii) Nuclear fusion (iii) Binding energy.

Type (IV) : Very Long Answer Type Questions: [05 Mark Each]35. What are the various postulates of Bohr's model of atom ? How could Bohr's model explain the existence of so

many lines in the spectrum of hydrogen ?

36. Write down electronic configuration of Fe3+ ion and answer the following :(i) What is the number of electrons having n + = 3 in it ?(ii) How many electrons in it have n = 3 and m = 0 ?(iii) How many electrons in it have = 1 ?(iv) What is the number of electrons in M-shell ?

37. Assign groups to the elements A, B, C, D and E starting from element X (group 2nd)

)AII(X AA B C D

What is relation between A and D ?What is relation between A, B and C?

38. Write down electronic configuration of chromium (Z=24) and indicate in it(i) number of sub-shells (ii) number of orbitals (iii) number of electrons in M-shell

39. Prove that En = – 13.6 × 2

2

nz

eV/atom for nth orbit in single electron species.


PART - I : SUBJECTIVE QUESTIONSSection (A) : Calculation related to nucleusA-1. Complete the following table :

Particle Atomic No. Mass No. No. of electrons No. of protons No. of neutrons

Sodium atom 11 --- --- --- 12

Aluminium ion --- 27 10 --- ---

Chloride ion --- --- 18 --- 18

Phosphorus atom --- 31 --- 15 ---

Cuprous ion --- --- 28 --- 35

A-2. If radius of the nucleus is 3.5 × 10–15 m then find the space or volume occupied by the nucleus.

A-3. The approximate radius of a H-atom is 0.05 nm, and that of proton is 1.5 × 10–15 m. Assuming both thehydrogen atom and the proton to be spherical, calculate fraction of the space in an atom of hydrogen that isoccupied by the nucleus.

A-4. (A) Find the radius of nucleus of an atom having atomic mass number equal to 125.(Take R0 = 1.3 × 10–15 m )

(B) Find the distance of closest approch when an particle is projected towards the nucleus of silver atomhaving speed v. (mass of particle = m, atomic number of Ag = 47)

Section (B) : Quantum theory of light & Photoelectric EfFectB-1. Calculate the energy of 100 photons if the wavelength of the light is 2000Å.

B-2. How many photones are emitted per second by a 5 mW laser operating at 620 nm?

B-3. The Vividh Bharati Station of All India Radio, Delhi broadcasts on a frequency of 1368 kHz (kilo hertz). Calculatethe wavelength and wave number of the electromagnetic radiation emitted by the transmitter.

B-4. One quantum is absorbed per gaseous molecule of Br2 for converting into Br atoms. If light absorbed haswavelength 5000 Å, calculate energy required in kJ/mol.

B-5. The eyes of a certain member of the reptile family pass a visual signal to the brain when the visual receptorsare struck by photons of wavelength 850 nm. If a total energy of 3.15 × 10–14 J is required to trip the signal,what is the minimum number of photons that must strike the receptor. (h = 6.6 × 10–34)

B-6. Two bulbs ‘A’ and ‘B’ emit red light and yellow light at 8000 Å and 4000Å respectively. The number of photonsemitted by both the bulbs per second is the same. If the red bulb is labelled as 100 watts, find the wattage ofthe yellow bulb.

B-7. The threshold frequency for the ejection of electrons from potassium metal is 5.3 × 1014 s–1. Will the photon ofa radiation having energy 3.3 × 10–19 J exhibit photoelectric effect ? (h = 6.626 × 10–34 Js)

B-8. If a light with frequency 4 × 1016 Hz emitted photoelectrons with double the maximum kinetic energy asare emitted by the light of frequency 2.5 × 1016 Hz from the same metal surface, then what is thethreshold frequency (0) of the metal ?

B-9. If the work function (w) of an arbitrary metal is 3.1 eV, find its threshold wavelength and the maximum kineticenergy of the electron emitted when radiation of 300 nm strike the metal surface. (Take hc = 12400 eVÅ)


Section (C) : Bohr ModelC-1. Which state of the triply ionized Beryllium (Be3+) has the same orbit radius as that of the ground state of

hydrogen atom ?

C-2. If the velocity of the electron in first orbit of H atom is 2.18 × 106 m/s,what is its value in third orbit ?

C-3. Consider Bohr's theory for hydrogen atom. The magnitude of angular momentum, orbit radius and velocity ofthe electron in nth energy state in a hydrogen atom are , r & v respectively. Find out the value of ' x', if productof v, r and (vr) is directly proportional to nx.

C-4. Find the ratio of the time period of 2nd Bohr orbit of He+ amd 4th Bohr orbit of Li2+.

C-5. Calculate the speed of an electron in the ground state of He+ ion. What fraction of speed of light is this value?How long does it take for the electron to complete one revolution around the nucleus. How many times doesthe electron travel around the nucleus in one second ?

C-6. Consider three electron jumps described below for the hydrogen atomx : n = 3 to n = 1y : n = 4 to n = 2z : n = 5 to n = 3

The photon emitted in which transition x, y or z will have shortest wavelength.

C-7. A hydrogen sample is prepared in a particular excited state. Photons of energy 2.55 eV get absorbed into thesample to take some of the electrons to a further excited state B. Find orbit numbers of the states A and B.Given the allowed energies of hydrogen atom :

E1 = –13.6 eV, E2 = –3.4 eV, E3 = –1.5 eV, E4 = –0.85 eV, E5 = –0.54 eV

C-8. When an electron falls from (n + 2) state to (n) state in a He+ ion the photon emitted has energy6.172 × 10–19 joules. What is the value of n.

C-9. A single electron ion has nuclear charge + Ze where Z is atomic number and e is electronic charge. It requires16.52 eV to excite the electron from the second Bohr orbit to third Bohr orbit. Find(a) The atomic number of element?(b) The energy required for transition of electron from first to third orbit?(c) Wavelength of photon required to remove electron from first Bohr orbit to infinity?(d) The kinetic energy of electron in first Bohr orbit?

C-10. The energy levels of hypothetical one electron atom are shown below. 0 eV ——— n = – 0.50 eV ——— n = 5– 1.45eV ——— n = 4– 3.08 eV ——— n = 3– 5.3 eV ——— n = 2– 15.6 eV ——— n = 1

(a) Find the ionisation potential of atom ?(b) Find the short wavelength limit of the series terminating at n = 2 ?(c) Find the wave no. of photon emitted for the transition made by the electron from third orbit to first orbit ?

C-11. The excitation energy of first excited state of a hydrogen like atom is 40.8 eV. Find the energy needed toremove the electron to form the ion.

Section (D) : SpectrumD-1. Calculate the two longest wavelengths of the radiation emitted when hydrogen atoms make transitions from

higher states to n = 2 state.

D-2. What electron transition in the He+ spectrum would have the same wavelength as the first Lyman transition ofhydrogen.

D-3. Calculate the frequency of light emitted for an electron transition from the sixth to second orbit of the hydrogenatom. In what region of the spectrum does this light occur?


D-4. At what atomic number would a transition from n = 2 to n = 1 energy level result in emission of photon of = 3 × 10–8 m.

D-5. Calculate the energy emitted when electrons of 1.0 g atom of hydrogen undergo transition giving the spectrallines of lowest energy in the visible region of its atomic spectra.RH = 1.1 × 107 mol–1, c = 3 × 108 m sec–1 and h = 6.62 × 10–34 J sec.

D-6. In a container a mixture is prepared by mixing of three samples of hydrogen, helium ion (He+) and lithium ion(Li2+). In sample all the hydrogen atoms are in 1st excited state and all the He+ ions are in third excited stateand all the Li2+ ions are in fifth excited state. Find the total number of spectral lines observed in the emissionspectrum of such a sample, when the electrons return back upto the ground state.

Section (E) : De broglie wavelength & uncertainty principleE-1. An electron in a H-atom in its ground state absorbs 1.5 times as much energy as the minimum required for its

escape (13.6 eV) from the atom. What is wavelength of the emitted electron.

E-2. Deduce the condition when the De-Broglie wavelength associated with an electron would be equal to thatassociated with a proton if a proton is 1836 times heavier than an electron.

E-3. An electron, practically at rest, is initially accelerated through a potential difference of 100 volts. It then has ade Broglie wavelength = 1 Å. It then get retarded through 19 volts and then has a wavelength 2 Å. A further

retardation through 32 volts changes the wavelength to 3Å, What is 1

23

?

E-4. If an electron having kinetic energy 2 eV is accelerated through the potential difference of 2 Volt. Then calculatethe wavelength associated with the electron

E-5. If uncertanity in the measurement of position and momentum are equal, calculate uncertainty in the measurementof velocity.

E-6. The uncertainty in position and velocity of the particle are 0.1 nm and 5.27×10–24 ms–1 respectively then findthe mass of the particle. (h = 6.625 × 10–34Js)

Section (F) : Quantum numbers & Electronic configurationF-1. How many unpaired electrons are there in Ni+2 ion if the atomic number of Ni is 28.

F-2. Write the electronic configuration of the element having atomic number 56.

F-3. Given below are the sets of quantum numbers for given orbitals. Name these orbitals.(a) n = 3 (b) n = 5 (c) n = 4 (d) n = 2 (e) n = 4 = 1 = 2 = 1 = 0 = 2

F-4. Point out the angular momentum of an electron in,(a) 4s orbital (b) 3p orbital (c) 4th orbit (according to Bohr model)

F-5. Which of the following sets of quantum numbers are impossible for electrons ? Explain why in each case.

F-6. Find the total spin and spin magnetic moment of following ion.(i) Fe+3 (ii) Cu+


Section (G) : Nuclear chemistryG-1. Calculate the loss in mass during the change :

Li73 + H1

1 2 He42 + 17.25 MeV

G-2. When 24Mg is bombared with neutron then a proton is ejected. Complete the equation and report the newelement formed.

G-3. Write equations for the following transformation :

(a) N147 (n, p) (b) ),n(K39

19

G-4. Explain with reason the nature of emitted particle by :

(a) Ca3820 (b) Ar35

18 (c) Ge8032 (d) Ca40

20

G-5. For the given series reaction in nth step, find out the number of produced neutrons & energy.

U23892 Ba + Kr + 30n

1 + Energy (E)

PART - II : OBJECTIVE QUESTIONS* Marked Questions may have more than one correct option.

Section (A) : Calculation related to nucleusA-1. The element having no neutron in the nucleus of its atom is

(A) Hydrogen (B) Nitrogen (C) Helium (D) Boron

A-2. The ratio of the "e/m" (specific charge) values of a electron and an -particle is -(A) 2 : 1 (B) 1 : 1 (C) 1 : 2 (D) None of these

A-3. The fraction of volume occupied by the nucleus with respect to the total volume of an atom is(A) 10–15 (B) 10–5 (C) 10–30 (D) 10–10

A-4.* Which of the following is iso-electronic with neon?(A) O2– (B) F– (C) Mg (D) Na

A-5. The approximate size of the nucleus of Ni6428 is :

(A) 3 fm (B) 4 fm (C) 5 fm (D) 2 fm

Section (B) : Quantum theory of light & Photoelectric EfFectB-1. The value of Planck’s constant is 6.63 × 10–34 Js. The velocity of light is 3 × 108 m/sec. Which value is closest

to the wavelength of a quantum of light with frequency of 8 × 1015 sec–1 ?(A) 5 × 10–18 m (B) 4 × 10–8 m (C) 3 × 107 m (D) 2 × 10–25 m

B-2. The MRI (magentic resonance imaging) body scanners used in hospitals operate with 400 MHz radio frequency.The wavelength corresponding to this radio frequency is(A) 0.75 m (B) 0.75 cm (C) 1.5 m (D) 2 cm

B-3. Photon of which light has maximum energy :(A) red (B) blue (C) violet (D) green

B-4. Electromagnetic radiations of wavelength 242 nm is just sufficient to ionise Sodium atom. Then the ionisationenergy of Sodium in kJ mole-1 is.(A) 494.65 (B) 400 (C) 247 (D) 600

B-5. Light of wavelength falls on metal having work function hc/0. Photoelectric effect will take place only if :(A) 0 (B) 20 (C) 0 (D) 0/2


B-6. A photon of energy h is absorbed by a free electron of a metal having work function w < h. Then :(A) The electron is sure to come out(B) The electron is sure to come out with a kinetic energy (h – w)(C) Either the electron does not come out or it comes with a kinetic energy (h – w)(D) It may come out with a kinetic energy less than (h – w)

B-7. A bulb of 40 W is producing a light of wavelength 620 nm with 80% of efficiency then the number of photonsemitted by the bulb in 20 seconds are (1eV = 1.6 × 10–19 J, hc = 12400 eV Å)(A) 2 × 1018 (B) 1018 (C) 1021 (D) 2 × 1021

Section (C) : Bohr ModelC-1. Correct order of radius of the st orbit of H, He+, Li2+, Be3+ is :

(A) H > He+> Li2+ > Be3+ (B) Be3+ > Li2+> He+ > H(C) He+ > Be3+ > Li2+ > H (D) He+ > H > Li2+ > Be3+

C-2. What is likely to be orbit number for a circular orbit of diameter 20 nm of the hydrogen atom :(A) 10 (B) 14 (C) 12 (D) 16

C-3. Which is the correct relationship :(A) E1 of H = 1/2 E2 of He+ = 1/3 E3 of Li2+ = 1/4 E4 of Be3+

(B) E1(H) = E2(He+) = E3 (Li2+) = E4(Be3+)(C) E1(H) = 2E2(He+) = 3E3 (Li2+) = 4E4(Be3+)(D) No relation

C-4. If the value of En = – 78.4 kcal/mole, the order of the orbit in hydrogen atom is :(A) 4 (B) 3 (C) 2 (D) 1

C-5. If velocity of an electron in orbit of H atom is V, what will be the velocity of electron in 3rd orbit of Li+2

(A) V (B) V/3 (C) 3 V (D) 9 V

C-6. In a certain electronic transition in the hydrogen atoms from an initial state (1) to a final state (2), the differencein the orbital radius (r1 – r2) is 24 times the first Bohr radius. Identify the transition.(A) 5 1 (B) 25 1 (C) 8 3 (D) 6 5

C-7. The species which has its fifth ionisation potential equal to 340 V is(A) B+ (B) C+ (C) B (D) C

C-8. Match the following(a) Energy of ground state of He+ (i) + 6.04 eV(b) Potential energy of orbit of H-atom (ii) –27.2 eV(c) Kinetic energy of excited state of He+ (iii) 54.4 V(d) Ionisation potential of He+ (iv) – 54.4 eV(A) A – (i), B – (ii), C – (iii), D – (iv) (B) A – (iv), B – (iii), C – (ii), D – (i)(C) A – (iv), B – (ii), C – (i), D – (iii) (D) A – (ii), B – (iii), C – (i), D – (iv)

C-9. S1 : Bohr model is applicable for Be2+ ion.S2 : Total energy coming out of any light source is integral multiple of energy of one photon.S3 : Number of waves present in unit length is wave number.S4 : e/m ratio in cathode ray experiment is independent of the nature of the gas.(A) F F T T (B) T T F F (C) F T T T (D) T F F F

C-10. S1 : Potential energy of the two opposite charge system increases with the decrease in distance.S2 : When an electron make transition from higher orbit to lower orbit it's kinetic energy increases.S3 : When an electron make transtition from lower energy to higher energy state its potential energy increases.S4 : 11eV photon can free an electron from the 1st excited state of He+ -ion.(A) T T T T (B) F T T F (C) T F F T (D) F F F F


C-11.* Choose the correct relations on the basis of Bohr’s theory.

(A) Velocity of electron n1

(B) Frequency of revolution 3n1

(C) Radius of orbit n2 Z (D) Electrostatic force on electron 4n1

Section (D) : SpectrumD-1. The energy of hydrogen atom in its ground state is –13.6 eV. The energy of the level corresponding to n = 5 is:

(A) –0.54 eV (B) –5.40 eV (C) –0.85 eV (D) –2.72 eV

D-2. Total no. of lines in Lyman series of H spectrum will be (where n = no. of orbits)(A) n (B) n – 1 (C) n – 2 (D) n (n + 1)

D-3. The wavelength of a spectral line for an electronic transition is inversely proportional to :(A) number of electrons undergoing transition(B) the nuclear charge of the atom(C) the velocity of an electron undergoing transition(D) the difference in the energy involved in the transition

D-4.* The spectrum of He+ is expected to be similar to that of :(A) Li2+ (B) He (C) H (D) Na

D-5. No. of visible lines when an electron returns from 5th orbit upto ground state in H spectrum :(A) 5 (B) 4 (C) 3 (D) 10

D-6. In a sample of H-atom electrons make transition from 5th excited state upto ground state, producing all possibletypes of photons, then number of lines in infrared region are(A) 4 (B) 5 (C) 6 (D) 3

D-7. Suppose that a hypothetical atom gives a red, green, blue and violet line spectrum . Which jump according tofigure would give off the red spectral line.

(A) 3 1 (B) 2 1 (C) 4 1 (D) 3 2

D-8. The difference between the wave number of 1st line of Balmer series and last line of paschen series for Li2+ ionis :

(A) 36R

(B) 36R5

(C) 4R (D) 4R

Section (E) : De broglie wavelength & uncertainty principleE-1. What possibly can be the ratio of the de Broglie wavelengths for two electrons each having zero initial energy

and accelerated through 50 volts and 200 volts ?(A) 3 : 10 (B) 10 : 3 (C) 1 : 2 (D) 2 : 1

E-2. In H-atom, if ‘x’ is the radius of the first Bohr orbit, de Broglie wavelength of an electron in 3rd orbit is :

(A) 3 x (B) 6 x (C) 2x9

(D) 2x

E-3. The approximate wavelength associated with a gold-ball weighing 200 g and moving at a speed of 5 m/hr is ofthe order of(A) 10–1 m (B) 10–20 m (C) 10–30 m (D) 10–40 m


E-4. An -particle is accelerated through a potential difference of V volts from rest. The de-Broglie’s wavelengthassociated with it is

(A) ÅV

150(B) Å

V286.0

(C) ÅV

101.0(D) Å

V983.0

E-5. de-Broglie wavelength of electron in second orbit of Li2+ ion will be equal to de-Broglie of wavelength of electronin(A) n = 3 of H-atom (B) n = 4 of C5+ ion (C) n = 6 of Be3+ ion (D) n = 3 of He+ ion

E-6. The wavelength of a charged particle ________the square root of the potential difference through which it isaccelerated :(A) is inversely proportional to (B) is directly proportional to(C) is independent of (D) is unrelated with

E-7. The uncertainty in the momentum of an electron is 1.0 × 10–5 kg m s–1 . The uncertainty in its position will be:(h = 6.626 × 10–34 Js)(A) 1.05 × 10–28 m (B) 1.05 × 10–26 m (C) 5.27 × 10–30 m (D) 5.25 × 10–28 m

Section (F) : Quantum numbers & Electronic configurationF-1. The orbital with zero orbital angular momentum is :

(A) s (B) p (C) d (D) f

F-2. Which of the following is electronic configuration of Cu2+ (Z = 29) ?(A) [Ar]4s1 3d8 (B) [Ar]4s2 3d10 4p1 (C) [Ar]4s1 3d10 (D) [Ar] 3d9

F-3. Spin magnetic moment of Xn+ (Z = 26) is 24 B.M. Hence number of unpaired electrons and value of nrespectively are :(A) 4, 2 (B) 2, 4 (C) 3, 1 (D) 0, 2

F-4. Which of the following ions has the maximum number of unpaired d-electrons?(A) Zn2+ (B) Fe2+ (C) Ni3+ (D) Cu+

F-5. The total spin resulting from a d7 configuration is :(A) 1 (B) 2 (C) 5/2 (D) 3/2

F-6. Given is the electronic configuration of element X :K L M N2 8 11 2

The number of electrons present with = 2 in an atom of element X is :(A) 3 (B) 6 (C) 5 (D) 4

F-7. Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum numbers, = 1 and 2 are, respectively :(A) 16 and 5 (B) 12 and 5 (C) 16 and 4 (D) 12 and 4

F-8. The orbital angular momentum of an electron in 2s-orbital is :

(A) + 2h

21

(B) zero (C) 2h

(D) 2h2

F-9. The possible value of and m for the last electron in the Cl– ion are :(A) 1 and 2 (B) 2 and + 1 (C) 3 and – 1 (D) 1 and – 1

F-10. For an electron, with n = 3 has only one radial node. The orbital angular momentum of the electron will be

(A) 0 (B) 6 2h

(C) 2 2h

(D) 3

2

h

F-11. The possible set of quantum no. for the unpaired electron of chlorine is :n m n m

(A) 2 1 0 (B) 2 1 1(C) 3 1 1 (D) 3 0 0


F-12.* Which of the following statement(s) is (are) correct?(A) The electronic configuration of Cr is [Ar] (3d)5 (4s)1. (Atomic number of Cr = 24)(B) The magnetic quantum number may have negative values.(C) In silver atom, 23 electrons have a spin of one type and 24 of the opposite type. (Atomic number of Ag = 47)(D) None of these

Section (G) : Quantum mechanical model of atomG-1 The correct time independent Schrödinger’s wave equation for an electron with E as total energy and V as

potential energy is :

(A) 0)VE(mh8

zyx 2

2

2

2

2

2

2

2

(B) 0)VE(

hm8

zyx 22

2

2

2

2

2

(C) 0)VE(h

m8zyx 2

2

2

2

2

2

2

2

(D) 0)VE(

hm8

zyx

2

2

2

2

2

2

2

G-2 The maximum radial probability in 1s-orbital occurs at a distance when : [ r0 = Bohr radius ]

(A) r = r0 (B) r = 2r0 (C) r = 2r0 (D) 2 r = 2

r0

G-3 Consider following figure A and B indicating distribution of charge density (electron probability 2) with dis-tance r.

Select the correct statement :(A) A and B both are for 1s (B) A and B both are for 2s(C) A is for 2s, B is for 1s (D) A is for 1s, B is for 2s

G-4 The maximum probability of finding electron in the dxy orbital is:(A) Along the x-axis (B) Along the y-axis(C) At an angle of 450 from the x and y axis (D) At an angle of 900 from the x and y axis.

G-5 3py orbital has..........nodal plane :(A) XY (B) YZ (C) ZX (D) All of these

G-6 A 3p-orbital has(A) Two non-spherical nodes (B) Two spherical nodes(C) One spherical and one non spherical nodes (D) One spherical and two non spherical nodes

G-7 According to Schrodinger model nature of electron in an atom is as :(A) Particle only (B) Wave only(C) Particle & wave nature both simultaneously (D) Sometimes waves and sometimes particle

G-8 The radial distribution curve of 2s sublevel consists of x nodes, x is :(A) 1 (B) 3 (C) 2 (D) 0

G-9 2zd orbital has :(A) Two lobe along Z axis & a ring in X-Y plane (B) A lobe & a ring along Z axis(C) A lobe along Z axis and a ring in Y-Z plane (D) None of these


G-10 Consider the following statements :(a) Electron density in the XY plane in 22 yxd3

orbital is zero

(b) Electron density in the XY plane in 2zd3 orbital is zero.

(c) 2s orbital has one nodal surface(d) for 2pz orbital, YZ is the nodal plane.Which of these are incorrect statements :(A) a & c (B) b & c (C) Only b (D) a, b, d

Section (H) : Nuclear chemistry

H-1* Pickout the correct statements :(A) Negative -decay decreases the proportion of neutrons and increases the proportion of proton.(B) Positive -decay increases the proportion of neutrons and decreases the proportion of proton.(C) K-electron capture increases the proportion of neutrons and increases the proporiton of proton.(D) Positrons and electrons quickly unite to produce photons.

H-2 C116 on decay produces :(A) Positron (B) -particle (C) -particle (D) none of these

H-3* Which consists of charged particles of matter?(A) -particle (B) -particle (C) -rays (D) Anode rays

H-4 Co6027 is radioactive because :

(A) its atomic number is high (B) it has high p/n ratio(C) it has high n/p ratio (D) none of these

H-5 Which of the following isotopes is likely to be most stable?

(A) Zn7130 (B) Zn66

30 (C) Zn6430 (D) None of these

H-6 Consider -particles, -particles and -rays, each having an energy of 0.50 MeV. The increasing order ofpenetration power is :(A) < < (B) < < (C) < < (D) < <

H-7 Al2713 is a stable isotope. Al29

13 is expected to disintegrate by :

(A) -emission (B) -emission (C) positron emission (D) proton emission

H-8 Which of the following nuclear emission will generate an isotope :(A) -emission (B) neutron emission (C) -emission (D) positron emission

H-9 The total number of - and -particles given out during given nuclear transformation is :

U23892 Pb214

82(A) 2 (B) 4 (C) 6 (D) 8

PART - III : ASSERTION / REASONING

DIRECTIONS :Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct.(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.(C) Statement-1 is True, Statement-2 is False.(D) Statement-1 is False, Statement-2 is True.(E) Statement-1 and Statement-2 both are False.

1. Statement-1 : Specific charge of -particle is twice to that of proton.Statement-2 : Specific charge is given by e/m.


2. Statement-1 : For n = 3, may be 0, 1 and 2 and 'm' may be 0, ± 1 and ± 2.Statement-2 : For each value of n, there are 0 to (n – 1) possible values of ; for each value of , there are 0to ± values of m.

3. Statement-1 : The possible number of electrons in a subshell is (4 + 2)Statement-2 : The possible number of orientations of a sub-shell are (2 + 1)

4. Statement-1 : If the potential difference applied to an electron is made 4 times, the de Broglie wavelengthassociated is halved. Initial kinetic energy of electron was zero.Statement-2 : On making potential difference 4 times, velocity is doubled and hence is halved.

5. Statement-1 : Wave number of a spectral line for an electronic transition is quantised.Statement-2 : Wave number is directly proportional to the velocity of electron undergoing the transition.

PART - I : SUBJECTIVE QUESTIONS1. One mole of He+ ion is excited from ground state such that 50% ions are in 3rd energy level 25% in 2nd

energy level while remaining were in ground state. Calculate the total energy evolved when all the ionsreturn to the ground state.

2. A bulb emits light of wave length 6200 Å. The bulb is rated as 150 watt and 8% of the energy is emitted as light.How many photons are emitted by bulb per second?

3. The latent heat of fusion of ice is 330 J/g. Calculate the number of photons of radiation of frequency5 × 1013 s–1 to cause the melting of 1 mole of ice. Take h = 6.6 × 10–34 J.S.

4. On the basis of Heisenberg's uncertainty principle, show that the moving electron can not exist within thenucleus.

5. For a beam of radiations coming out from a sample of He+ ions the following energy level diagram was observed (The beam only had photons correspondingto these transitions). If the above beam of radiations is used to excite a sample ofH–atoms in ground state then find the wavelengths of the photons subsequentlyemitted by the H–atoms.

6. In a sample of hydrogen atom in ground state electrons make transition from ground state to a particularexcited state where path length is five times de-broglie wavelength, electrons make back transition to theground state producing all possible photons. If photon having 2nd highest energy of this sample can be usedto excite the electron in a particular excited state of Li2+ atom then find the final excited state of Li2+ atom.

7. A gas of identical H-like atom has some atoms in the lowest (ground) energy level A and some atoms in aparticular upper (excited) energy level B and there are no atoms in any other energy level. The atoms of thegas make transition to a higher energy level by absorbing monochromatic light of photon energy 2.7 eV.Subsequently, the atoms emit radiation of only six different photons energies. Some of the emitted photonshave energy 2.7 eV. Some have more and some have less than 2.7 eV.(a) Find the principal quantum number of initially excited level B.(b) Find the ionisation energy for the gas atoms.(c) Find the maximum and the minimum energies of the emitted photons.

8. Using Bohr's theory show that when n is very large, the frequency of radiation emitted by hydrogen atom dueto transition of electron from n to (n – 1) is equal to frequency of revolution of electron in its orbit.

9. Electrons in a sample of H-atoms make transition from state n = x to some lower excited state. The emissionspectrum from the sample is found to contain only the lines belonging to a particular series. If maximum

energy photon has an energy of 0.6375 ev, find the value of x. [Take 0.6375 eV = 43

× 0.85 eV]


10. An electron in Li2+ ion makes a transition from higher state n2 to lower state n1 = 6. The emitted photons is usedto ionize an electron in H–atom from 2nd excited state. The electron on leaving the H–atom has a de–Brogliewavelength = 12.016 Å. Find the value of n2.

Note : Use (12.016)2 = 116.13144150

Å = eVKE

150.

PART - II : OBJECTIVE QUESTIONSSingle choice type1. Which orbital is non-directional

(A) s (B) p (C) d (D) All

2. Uncertainty in position is twice the uncertainty in momentum. Uncertainty in velocity is :

(A) h

(B) h

m21

(C) m21

(D) 4

h

3. For which orbital angular probabili ty distribution is maximum at an angle of 45° to the axialdirection-

(A) 22 yxd (B) 2zd (C) dxy (D) Px

4. If n and are respectively the principal and azimuthal quantum numbers, then the expression for calculatingthe total number of electrons in any orbit is -

(A)

n

1

)12(2

(B)

1n

1

)12(2

(C)

1n

0

)12(2

(D)

1n

0

)12(2

5. If wavelength is equal to the distance travelled by the electron in one second, then -

(A) = ph

(B) = mh

(C) = ph

(D) = mh

6. The quantum numbers + 1/2 and – 1/2 for the electron spin represent -(A) Rotation of the electron in clockwise and anticlockwise direction respectively.(B) Rotation of the electron in anticlockwise and clockwise direction respectively.(C) Magnetic moment of the electron pointing up and down respectively,(D) Two quantum mechanical spin states which have no classical analogue.

7. Consider an electron in the nth orbit of a hydrogen atom in the Bohr model. The circumference of the orbit canbe expressed in terms of the de Broglie wavelength of the electron as :

(A) (0.529) n (B) n (C) (13.6) (D) n

8. A particle X moving with a certain velocity has a debroglie wave length of 1Å, If particle Y has a mass of 25%that of X and velocity 75% that of X, debroglies wave length of Y will be -(A) 3 Å (B) 5.33 Å (C) 6.88 Å (D) 48 Å

9. The wave number of electromagnetic radiation emitted during the transition of electron in between two levelsof Li2+ ion having sum of the principal quantum numbers 4 and difference is 2, will be : (RH = Rydberg constant)

(A) 3.5 RH (B) 4 RH (C) 8 RH (D) 98

RH

10. De Broglie wavelength of an electron after being accelerated by a potential difference of V volt from rest is

(A) = Åh3.12

(B) = ÅV

3.12(C) = Å

E3.12

(D) = Åm

3.12


11. What are the values of the orbital angular momentum of an electron in the orbitals 1s, 3s, 3d and 2p -

(A) 0, 0, 6 , 2 (B) 1, 1, 4 , 2 (C) 0, 1 6 , 3 (D) 0, 0 20 , 6

12. After np orbitals are filled, the next orbital filled will be :(A) (n + 1) s (B) (n + 2) p (C) (n + 1) d (D) (n + 2) s

13. The value of the spin magnetic moment of a particular ion is 2.83 Bohr magneton. The ion is :(A) Fe2+ (B) Ni2+ (C) Mn2+ (D) Co3+

14. In Bohr's model of the hydrogen atom the ratio between the period of revolution of an electron in the orbit of n= 1 to the period of the revolution of the electron in the orbit n = 2 is -(A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 1 : 8

15. In an atom, two electrons move round the nucleus in circular orbits of radii R and 4R. The ratio of the timetaken by them to complete one revolution is : (Consider Bohr model to be valid)(A) 1 : 4 (B) 4 : 1 (C) 1 : 8 (D) 8 : 1

16. Let 1 be the frequency of the series limit of the Lyman series, 2 be the frequency of the first line of the Lymanseries, and 3 be the frequency of the series limit of the Balmer series -(A) 1 – 2 = 3 (B) 2 – 1 = 3 (C) 3 = 1/2 (1 – 3) (D) 1 + 2 = 3

17. If first ionization potential of a hypothetical atom is 16 V, then the first excitation potential will be :(A) 10.2 V (B) 12 V (C) 14 V (D) 16 V

18. The angular momentum of an electron in a given orbit is J, Its kinetic energy will be :

(A) 2

2

mrJ

21

(B) r

Jv(C)

m2J2

(D) 2

J2

19. According to Bohr's model of hydrogen atom the electric current generated due to motion of electron in nth

orbit is:

(A) 22

422

hnemk4

(B) 22

522

hnemk4

(C) 522

22

emk4hn

(D) 33

522

hnemk4

20. Total number of electrons having n + = 3 in Cr (24) atom in its ground state is :(A) 8 (B) 10 (C) 12 (D) 6

21. The correct set of four quantum numbers for the valence electron of Rubidium (Z = 37) is

(A) n = 5, = 0, m = 0, s = + 21

(B) n = 5, = 1, m = 0, s = + 21

(C) n = 5, = 1, m = 1, s = + 21

(D) n = 6, = 0, m = 0, s = + 21

22. No. of visible lines when an electron returns from 5th orbit upto ground state in H spectrum :(A) 5 (B) 4 (C) 3 (D) 10

23. If the shortest wave length of Lyman series of H atom is x, then the wave length of the first line of Balmer seriesof H atom will be :(A) 9x/5 (B) 36x/5 (C) 5x/9 (D) 5x/36

24. Which of the given statement (s) is/are false.I. Orbital angular momentum of the electron having n = 5 and having value of the azimuthal quantum number

as lowest for this principle quantum number is h

.II. If n = 3, = 0, m = 0, for the last valence shell electron, then the possible atomic number must be 12 or 13.

III. Total spin of electrons for the atom 25Mn is ± 27

.IV. Spin magnetic moment of inert gas is 0 .

(A) I, II and III (B) II and III only (C) I and IV only (D) None of these


25. An electron in a hydrogen like atom makes transition from a state in which its de-Broglie wavelength is 1 to astate where its de-Broglie wavelength is 2 then wavelength of photon () generated will be

(A) = 1 – 2 (B) =

hmc4

22

21

22

21

(C) = 22

21

22

21

(D) =

hmc2

22

21

22

21

where m is mass of the electron, c is speed of light in vaccum.

26. Photon having energy equivalent to the binding energy of 4th state of He+ atom is used to eject an electronfrom the metal surface of work function 1.4 eV. If electrons are further accelerated through the potentialdifference of 4V then the minimum value of De–broglie wavelength associated with the electron is :(A) 1.1 Å (B) 5 Å (C) 9.15 Å (D) 11 Å

27. In a sample of H-atoms, electrons de-excite from a level ‘n’ to 1. The total number of lines belonging to Balmerseries are two. If the electrons are ionised from level ‘n’ by photons of energy 13 eV. Then the kinetic energy ofthe ejected photoelectrons will be(A) 12.15 eV (B) 11.49 eV (C) 12.46 eV (D) 12.63 eV

28. In case of 22 yxd

orbital

(A) Probability of finding the electron along x-axis is zero.(B) Probability of finding the electron along y-axis is zero.(C) Probability of finding the electron is maximum along x and y-axis.(D) Probability of finding the electron is zero in x-y plane

29. Change in angular momentum when an electron makes a transition corresponding to the 3rd line of the Balmerseries in Li2+ ion is

(A) 2h

(B) 2h2

(C) 2h3

(D) 2h4

30. An -particle has initial kinetic energy of 25 eV and it is accelerated through the potential difference of 150 volt.If a proton has initial kinetic energy of 25 eV and it is accelerated through the potential difference of 25 volt thenfind the approximate ratio of the final wavelengths associated with the proton and the -particle.(A) 5 (B) 4 (C) 3 (D) 2

More than one choice type31. A hydrogen - like atom has ground state binding energy 122.4 eV. Then :

(A) its atomic number is 3(B) a photon of 90 eV can excite it to a higher state(C) a 80 eV photon cannot excite it to a higher state(D) None

32. A sodium street light gives off yellow light that has a wavelength of 600 nm. Then

(For energy of a photon take E = )Å(ÅeV12400

)

(A) frequency of this light is 7× 1014 s–1 (B) frequency of this light is 5× 1014 s–1

(C) wavenumber of the light is 3 × 106 m–1 (D) energy of the photon is approximately 2.07 eV

33. The qualitative order of Debroglie wavelength for electron, proton and particle is e > P > if(A) If kinetic energy is same for all particles(B) If the accelerating potential difference 'V' is same for all the particles (from rest)(C) If velocities are same for all particles(D) None of the above


34. If there are only two H-atoms, each is in 3rd excited state then :(A) Maximum number of different photons emitted is 4.(B) Maximum number of different photons emitted is 3.(C) Minimum number of different photons emitted is 1.(D) Minimum number of different photons emitted is 2.

35. Which of the following statements is/are correct for an electron of quantum numbers n = 4 and m = 2?(A) The value of may be 2. (B) The value of may be 3.(C) The value of s may be +1/2. (D) The value of may be 0, 1, 2, 3.

PART - III : MATCH THE COLUMN & COMPREHENSION1. Column I Column II

(i) Aufbau principle (a) Line spectrum in visible region(ii) de broglie (b) Maximum multiplicity of electron(iii) Angular momentum (c) Photon(iv) Hund’s rule (d) = h/(mv)(v) Balmer series (e) Electronic configuration(vi) Planck’s law (f) mvr

2. Column I Column II(i) Cathode rays (a) Helium nuclei(ii) Dumb-bell (b) Uncertainty principle(iii) Alpha particles (c) Electromagnetic radiation(iv) Moseley (d) p-orbital(v) Heisenberg (e) Atomic number(vi) X-rays (f) Electrons

3. Frequency = f, Time period = T, Energy of nth orbit = En , radius of nth orbit = rn , Atomic number = Z,Orbit number = nColumn I Column II(i) f (a) n3

(ii) T (b) Z2

(iii) En (c) 2n1

(iv) nr1

(d) Z

4. Column I Column II(i) Lyman series (a) maximum number of spectral line observed = 6(ii) Balmer series (b) maximum number of spectral line observed = 2

(iii) In a sample of H-atom (c) 2nd line has wave number 9R8

for 5 upto 2 transition

(iv) In a single isolated H-atom (d) 2nd line has wave number 16R3

for 3 upto 1 transition(e) Total number of spectral line is 10.


PART - IV : COMPREHENSIONRead the following passage carefully and answer the questions.

Comprehension # 1In the photoelectric effect the electrons are emitted instantaneously from a given metal plate, when it is irradiatedwith radiation of frequency equal to or greater than some minimum frequency, called the threshold frequency.According to planck's idea, light may be considered to be made up of discrete particles called photons . Eachphoton carries energy equal to h. When this photon collides with the electron of the metal, the electronacquires energy equal to the energy of the photon. Thus the energy of the emitted electron is given by :

h = K.Emaximum + P. E. = 21

mu2 + PE

If the incident radiation is of threshold frequency the electron will be emittedwithout any kinetic energy i.e. h0 = PE

21

mu2 = h – h0

A plot of kinetic energy of the emitted electron versus frequency of the incidentradiation yields a straight line given as

1. A beam of white light is dispersed into its wavelength components by a Quartz prism and falls on a thin sheetof potassium metal. What is the correct decreasing order of kinetic energy of maximum the electron emittedby the different light component.(A) blue > green > orange > yellow (B) violet > blue > orange > red(C) yellow > green > blue > violet (D) orange > yellow > blue > violet

2. A laser producing monochromatic light is used to eject electron from the sheet of gold having threshold frequency6.15 x 1014 s–1 which of the following incident radiation will be suitable for the ejection of electron :(A) 1.5 moles of photons having frequency 3.05 × 1014 s–1

(B) 0.5 moles of photon of frequency 12.3 × 1012 s–1

(C) One photon with frequency 5.16 × 1015 s–1

(D) All of the above

3. The number of photoelectrons emitted depends upon :(A) The intensity of the incident radiation(B) The frequency of the incident radiation(C) The product of intensity and frequency of incident radiation(D) None of these

Comprehension # 2The only electron in the hydrogen atom resides under ordinary conditions on the first orbit. When energy is supplied, the electron moves to higherenergy orbit depending on the amount of energy absorbed. When thiselectron returns to any of the lower orbits, it emits energy. Lyman seriesis formed when the electron returns to the lowest orbit while Balmerseries is formed when the electron returns to second orbit. Similarly,Paschen, Brackett and Pfund series are formed when electron returnsto the third, fourth and fifth orbits from higher energy orbits respectively(as shown in figure)Maximum number of lines produced when electrons jump from nth level

to ground level is equal to 2

)1n(n .

For example, in the case of n = 4, number of lines produced is 6. (4 3, 4 2, 4 1, 3 2, 3 1, 2 1).When an electron returns from n2 to n1 state, the number of lines in the spectrum will be equal to

2)1nn)(nn( 1212


If the electron comes back from energy level having energy E2 to energy level having energy E1, then thedifference may be expressed in terms of energy of photon as :

E2 – E1 = E ,E

hc

, E = h ( - frequency)

Since h and c are constants, E corresponds to definite energy; thus each transition from one energy level toanother will produce a light of definite wavelength. This is actually observed as a line in the spectrum ofhydrogen atom.

Wave number of line is given by the formula

2

221

2

n1

n1RZ .

where R is a Rydberg constant (R = 1.1 × 107 m–1)(i) First line of a series : It is called ‘line of longest wavelength’ or ‘line of shortest energy’.(ii) Series limit or last line of a series : It is the line of shortest wavelength or line of highest energy.

4. Last line of Brackett series for H–atom has wavelength 1 Å and 2nd line of lyman series has wavelength 2 Å,

then :

(A) 1

128 =

2

9 (B)

1

16 =

2

9 (C)

1

4 =

2

1 (D)

1

128 =

2

8

5. Consider the following statements1. Spectral lines of He+ ion belonging to Balmer series are not in visible range.2. In the balmer series of H-atom maximum lines are in ultra violet region.3. 2nd line of lyman series of He+ ion has energy 48.4 eVThe above statements 1, 2, 3 respectively are (T = True, F = False)(A) T F F (B) F T T (C) T F T (D) T T T

6. Wave number of the first line of Paschen series in Be3+ ion is

(A) 16R7

(B) 144

R7(C) 9

R7(D)

144R

Comprehension # 3

de Broglie proposed dual nature for electron by putting his famous equation = mvh

. Later on Heisenberg

proposed uncertainty principle as p. x 4h

. On the contrary, particle nature of electron was established on

the basis of photoelectric effect. When a photon strikes the metal surface, it gives up its energy to the electron.Part of this energy (say W) is used by the electrons to escape from the metal and the remaining energyimparts kinetic energy (1/2 mv2) to the ejected photoelectron. The potential applied on the surface to reducethe velocity of photoelectron to zero is known as stopping potential.

7. Uncertainity in the position of an electron (mass 9.1 10–31 kg) moving with a velocity 300 ms–1, accurate upto

0.001% will be : (em2

= 5.8 10–5)

(A) 19.2 10–2 m (B) 5.76 10–2 m (C) 3.84 10–2 m (D) 1.92 10–2 m

8. When a beam of photons of a particular energy was incident on a surface of a particular pure metal havingwork function = (40 eV), some emitted photoelectrons had stopping potential equal to 22 V, some had 12 Vand rest had lower values. Calculate the wavelength of incident photons assuming that at least one photoelectronis ejected with maximum possible kinetic energy. .(A) 310 Å (B) 298 Å (C) 238 Å (D) 200 Å

9. The circumference of third orbit of a single electron species is 3 nm. What may be the approximate wavelengthof the photon required to just ionize electron from this orbit.(A) 91.1 nm (B) 364.7 nm (C) 821 nm (D) 205 nm


Comprehension # 4Quantum numbers are assigned to get complete information of electrons regarding their energy, angularmomentum, spectral lines etc. Four quantum numbers are known i.e. principal quantum numbers which tellthe distance of electron from nucleus, energy of electron in a particular shell and its angular momentum.Azimuthal quantum number tells about the subshells in a given shell and of course shape of orbital. Magneticquantum number deals with study of orientations or degeneracy of a subshell.

Spin quantum number which defines the spin of electron designated as +21

or –21

represented by and

respectively. Electron are filled in orbitals following Aufbau rule. Pauli's exclusion principal and Hund's rule ofmaximum multiplicity. On the basis of this answer the following questions.

10. Two unpaired electrons present in carbon atom are different with respect to their(A) Principle quantum number (B) Azimuthul quantum number(C) Magnetic quantum number (D) Spin quantum number

11. Number of electron having the quantum numbers n = 4, = 0, s = 21

in Zn+2 ion is/are :

(A) 1 (B) 0 (C) 2 (D) 5

12. Spin angular momentum for unpaired electron in sodium (Atomic No. = 11) is

(A) 23

(B) 0.866 h/2 (C) –23

2h

(D) None of these

Comprehension # 5Read the following passage carefully and answer the following objective questions based on thepassageAfter the failure of Bohr atomic theory but its ability to explain the atomic spectra a need was felt for the newmodel that could incorporate, the concept of stationary orbit, de Broglie concept, Heisenberg uncertaintyprinciple. The concept that in corporate above facts is called quantum mechanics of the atomic model wavemechanical model. It includes set of quantum numbers and |2|a mathematical expression of the probability offinding an electron at all points in space. This probability function is the best indication available of how theelectron behaves, for as a consequence of the Uncertainty Principle, the amount we can know about theelectron is limited. While quantum mechanics can tell us the exact probability of finding an electron at any twoparticular points, it does not tell us how the electron moves from one of these points to the other. Thus the ideaof an electron orbit is lost; it is replaced with a description of where the electron is most likely to be found. Thistotal picture of the probability of finding an electron at various points in space is called an orbital.

There are various types of orbitals possible, each corresponding to one of the possible combinations of quan-tum numbers. These orbitals are classified according to the value of n and l associated with them. In order toavoid confusion over the use of two numbers, the numerical values of l are replaced by letters; electrons inorbitals with l = 0 are called s-electrons those occupying orbitals for which l = 1 are p-electrons and those forwhich l = 2 are called d-electrons. The numerical and alphabetical correspondences are summarized in table.Using the alphabetical notation for l, we would say that in the ground state of hydrogen atom (n = 1, l = 0) wehave a 1s-electron, or that the electron moves in a 1s-orbital. The relation of the spherical polar co-ordinatesr, and to Cartesian coordinates x, y and z. To make the concept of an orbital more meaningful, it is helpfulto examine the actual solution of the wave function for the one-electron atom. Because of the sphericalsymmetry of the atom, the wave functions are most simply expressed in terms of a spherical polar-coordinatesystem, shown in fig., which has its orbit at the nucleus. It is found that the wave functions can be expressed


as the product of two functions, one of which (the “angular part” X) depends only the angle and , the otherof which (the “radial part” R) depends only on the distance from the nucleus. Thus we have

(r, , ) = R(r) X (, )Angular and radial parts of hydrogen atom wave functions

Angular part X() Radial part Rn, (r)

X(s) = 2/1

41

R(1s) = 2 2/

2/3

0e

az

X(px) = 2/1

43

sin cos R(2s) = 221

2/3

0az

(2 – ) 2/e

X(py) = 2/1

43

sin sin R(2p) = 621

2/3

0az

2/e

X(pz) = 2/1

43

cos

X(dz2) =

2/1

165

(3 cos2 – 1)

X(dxz) = 2/1

415

sin coscos R(3s) = 391

2/3

0az

2/2 e)66(

X(dyz) = 2/1

415

sin cossin R(3p) = 691

2/3

0az

2/e)4(

X(dx2 – y2) = 2/1

415

sin2 cos2 R(3d) = 3091

2/3

0az

2/2e

X(dxy) = 2/1

415

sin2 sin 2

0naZr2

22

2

0 me4ha

This factorization helps us to visualize the wave function, since it allows us to consider the angular and radialdependences separately. It contains the expression for the angular and radial parts of the one electron atomwave function. Note that the angular part of the wave function for an s-orbital it always the same, (1/4)1/2,regardless of principal quantum number. It is also true that the angular dependence of the p-orbitals and of thed-orbitals is independent of principle quantum number. Thus all orbitals of a given types (s, p, or d) have thesame angular behaviour The table shows, however, that the radial part of the wave function depends both onthe principal quantum number n and on the angular momentum quantum number l.To find the wave function for a particular state, we simply multiply the appropriate angular and radial partstogether called normalized wave function.The probability of finding an electron at a point within an atom is proportional to the square of orbital wavefunction, i.e., 2 at that point. Thus, 2 is known as probability density and alwyas a positive quantity.2 dV (or 2.4r2dr). represents the probability for finding electron in a small volume dV surrounding thenucleus.


13. The electron probability density for 1s-orbital is best represented by the relation

(A) 0ar2/3

0e

aZ

21

(B) 0a

zr23

0e

aZ1

(C) 0ar2/3

0e

aZ1

(D) 0azr23

0e

aZ2

14. The wave function () of 2s-orbital is given by :

2s = 32

1 2/3

0a1

0a2/r

0e

ar–2

, At r = r0, radial node is formed.

Then which of the following is correct :(A) r0 = a0 (B) r0 = 2a0 (C) r0 = 3a0 (D) None of these

15. The angular wave function of which orbital will not disturb by the variation with azimuthal angle only(A) 1s and 2s (B) 2pz and 2dz

2 (C) 2px and 3dz2 (D) 2px and 2s

PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS)* Marked Questions may have more than one correct option.

1. Calculate the energy required to excite one litre Hydrogen gas at 1 atm and 298 K to first excited state ofatomic hydrogen. The energy for the dissociation of H-H bond is 436 kJ mol-1. [JEE 2000, 4/135]

2. The orbit having Bohr radius equal to 1st Bohr orbit of H–atom is [JEE 2004, 3/144](A) n = 2 of He+ (B) n = 2 of B+4 (C) n = 3 of Li+2 (D) n = 2 of Be+3

3. (a) The wave function of an electron in 2s orbital in hydrogen atom is given below :

2s = 2/1)2(41

2/3

0az

0ar2 exp (– r/2a0)

where a0 is the Bohr radius. This wave function has a radial node at r = r0. Express r0 in terms ofa0. [JEE 2004, 4/60]

(b) Calculate the wavelength of a ball of mass 100 g moving with a velocity of 100 ms–1.

(c) 92X238

6–

8– Y. Find out atomic number, mass number of Y and identify it.

4. (a) Using Bohr’s model for hydrogen atom, find the speed of electron in the first orbit if the Bohr’s radius isa0 = 0.529 × 10–10 m. Find deBroglie wavelength of the electron also. [JEE 2005 ,4/144]

(b) Find the orbital angular momentum of electron if it is in 2p orbital of H in terms of 2

h.

5. According to Bohr’s theory,En = Total energy , Kn = Kinetic energyVn = Potential energy , rn = Radius of nth orbit

Match the following: [JEE 2006 ,6/184]Column I Column II(A) Vn / Kn = ? (p) 0(B) If radius of nth orbit En

x , x = ? (q) – 1(C) Angular momentum in lowest orbital (r) – 2

(D) ?y,Zr1 y

n (s) 1


Paragraph for Question Nos. 6 to 8The hydrogen-like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbinglight the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to theground state energy of the hydrogen atom.

6. The state S1 is : [JEE 2010, 3/163](A) 1s (B) 2s (C) 2p (D) 3s

7. Energy of the state S1 in units of the hydrogen atom ground state energy is : [JEE 2010, 3/163](A) 0.75 (B) 1.50 (C) 2.25 (D) 4.50

8. The orbital angular momentum quantum number of the state S2 is : [JEE 2010, 3/163](A) 0 (B) 1 (C) 2 (D) 3

9. The work function () of some metals is listed below. The number of metals which will show photoelectriceffect when light of 300 nm wavelength falls on the metal is [JEE 2011, 4/180]

Metal Li Na K Mg Cu Ag Fe Pt W

(eV) 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75

10. The maximum number of electrons that can have principal quantum number, n = 3, and spin quantum number,ms = –1/2, is [JEE 2011, 4/180]

11. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius] :[JEE 2012, 3/136]

(A) 20

2

2

ma4h

(B) 20

2

2

ma16h (C) 2

02

2

ma32h (D) 2

02

2

ma64h

12. Bombardment of aluminum by -particle leads to its artificial disintegration in two ways, (I) and (ii) as shown.Products X, Y and Z respectively are,

[JEE 2011, 3/180]

(A) proton, neutron, positron (B) neutron, positron, proton(C) proton, positron, neutron (D) positron, proton, neutron

13. The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes anuclear reaction yielding element X as shown below. To which group, element X belongs in the periodic table?

XH2n6HCu 11

10

11

6329 [JEE 2012, 4/136]

PART - II : AIEEE PROBLEMS (PREVIOUS YEARS)1. Which of the following ions has the maximum magnetic moment? [AIEEE 2002]

(A) Mn+2 (B) Fe+2 (C) Ti+2 (D) Cr+2.

2. Energy of H-atom in the ground state is –13.6 eV, hence energy in the second excited state is[AIEEE 2002]

(A) – 6.8 eV (B) – 3.4 eV (C) – 1.51 eV (D) – 4.53 eV

3. Uncertainity in position of a particle of 25 g in space is 10–15 m. Hence, Uncertainity in velocity (in m.sec–1)is : (plank’s constant, h = 6.6 10–34 Js) [AIEEE 2002](A) 2.1 10–18 (B) 2.1 10–34 (C) 0.5 10–34 (D) 5.0 10–24


4. The de-Broglie wavelength of a tennis ball of mass 60 g moving with a velocity of 10 m/s is approximately(planck’s constant, h = 6.63 10–34 J-s) [AIEEE 2003](A) 10–33 m (B) 10–31 m (C) 10–16 m (D) 10–25 m

5. In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of thefollowing inner-orbit jumps of the electron for Bohr orbits in an atom of hydrogen ? [AIEEE 2003](A) 3 2 (B) 5 2 (C) 4 1 (D) 2 5

6. The numbers of d-electrons retained in Fe2+ (atomic number Fe = 26) ion is [AIEEE 2003](A) 3 (B) 4 (C) 5 (D) 6

7. The orbital angular momentum for an electron revolving in an orbit is given by )1( 2

h. This momentum

for an s-electron will be given by [AIEEE 2003]

(A) +21

. 2h

(B) Zero (C) 2h

(D) 2 . 2h

8. The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state1, would be (Rydberg constant = 1.097 107 m–1) [AIEEE 2004](A) 91 nm (B) 192 nm (C) 406 (D) 9.1 10–6 nm

9. Which of the following set a of quantum numbers is correct for an electron in 4f orbital? [AIEEE 2004](A) n = 4, l =3, m = +4, s = +1/2 (B) n = 4, l = 4, m = –4, s = –1/2(C) n = 4, l = 3, m = +1, s = +1/2 (D) n = 3, l=2, m =–2, s = +1/2

10. Consider the ground state of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum numbers, = 1 and 2 are, respectively [AIEEE 2004](A) 12 and 4 (B) 12 and 5 (C) 16 and 4 (D) 16 and 5

11. Which of the following statements in relation to the hydrogen atom is correct ? [AIEEE 2005](A) 3s, 3p and 3d orbitals all have the same energy(B) 3s and 3p orbitals are of lower energy than 3d orbital(C) 3p orbital is lower in energy than 3d orbital(D) 3s orbital is lower in energy than 3p orbital

12. In a multi-electron atom, which of the following orbitals described by the three quantum numbers will havethe same energy in the absence of magnetic and electric field ? [AIEEE 2005](i) n = 1, l = 0, m = 0 (ii) n =2, l = 0, m = 0 (iii) n = 2, l = 1, m = 1 (iv) n = 3, l = 2, m =1(v) n = 3, l = 2, m = 0(A) (iv) and (v) (B) (iii) and (iv) (C) (ii) and (iii) (D) (i) and (ii)

13. Uncertainity in the position of an electron (mass = 9.1 10–31 Kg) moving with a velocity 300 m.sec–1, Accurateupto 0.001%, will be : (h = 6.63 10–34 J-s) [AIEEE 2006](A) 19.2 10–2 m (B) 5.76 10–2 m (C) 1.92 10–2 m (D) 3.84 10–2 m

14. According to Bohr’s theory, the angular momentum to an electron in 5th orbit is : [AIEEE 2006]

(A) 25 h

(B) 1.0 h

(C) 10 h

(D) 2.5 h

15. The ‘spin-only’ magnetic moment [in units of Bohr magneton ()] of Ni2+ in aqueous solution would be (Atomicnumber : Ni = 28) [AIEEE 2006](A) 2.84 (B) 4.90 (C) 0 (D) 1.73

16. Which of the following nuclear reactions will generate an isotope ? [AIEEE 2007, 3/120](A) Neutron particle emission (B) Positron emission(C) -particle emission (D) -particle emission


17. The ionisation enthalpy of hydrogen atom is 1.312 106 J mol–1. The energy required to excite the electronin the atom from n1 = 1 to n2 = 2 is [AIEEE 2008, 3/105](A) 8.51 105 J mol–1 (B) 6.56 105 J mol–1

(C) 7.56 105 J mol–1 (D) 9.84 105 J mol–1

18. Which of the following set of quantum numbers represents the highest energy of an atom ?

(A) n = 3, l = 0, m = 0, s = + 21

(B) n = 3, l = 1, m =1, s = +21

[AIEEE 2008, 3/105]

(C) n = 3, l = 2, m = 1, s = + 21

(D) n = 4, l = 0, m = 0, s = + 21

19. The energy required to break one mole of Cl – Cl bonds in Cl2 is 242 kJ mol–1 . The longest wavelength of lightcapable of breaking a single Cl – Cl bond is [AIEEE 2010, 4/144](c = 3 × 108 m s–1 and NA = 6.02 × 1023 mol–1)(A) 594 nm (B) 640 nm (C) 700 nm (D) 494 nm

20. Ionisation energy of He+ is [AIEEE 2010, 4/144]19.6 × 10–18 J atom–1 . The energy of the first stationary state (n = 1) of Li2+ is :(A) 4.41 × 10–16 J atom–1 (B) – 4.41 × 10–17 J atom–1

(C) – 2.2 × 10–15 J atom–1 (D) 8.82 × 10–17 J atom–1

21. A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the otheris at : [AIEEE 2011, 4/120](A) 1035 nm (B) 325 nm (C) 743 nm (D) 518 nm

22. The frequency of light emitted for the transition n = 4 to n = 2 of He+ is equal to the transition in H atomcorresponding to which of the following? [AIEEE 2011, 4/120](A) n = 2 to n = 1 (B) n = 3 to n = 2 (C) n = 4 to n = 3 (D) n = 3 to n = 1

23. The electrons identified by quantum numbers n and : [AIEEE 2012, 4/120](a) n = 4, = 1 (b) n = 4, = 0(c) n = 3, = 2 (d) n = 3, = 1can be placed in order of increasing energy as :(A) (c) < (d) < (b) < (a) (B) (d) < (b) < (c) < (a)(C) (b) < (d) < (a) < (c) (D) (a) < (c) < (b) < (d)


BOARD LEVEL EXERCISE : HINT & SOLUTIONS

1. (i) P3115

Number of electrons = Atomic number = 15Number of protons = Atomic number = 15Number of neutrons = Mass number – Atomic number = 31 – 15 = 16

(ii) 33115P

P3– = P + 3e–

Thus, phosphide ion contains same number of protons and neutrons but contains three electrons more thanphosphorus atom.

Number of electrons = 15 + 3 = 18Number of protons= 15Number of neutrons = 31 – 15 = 16

2. The rest mass of photon is zero.

3. E1 = – 2

2

)1()2(6.13

= – 54.4 ev / atom.

4. 1 mole of NH3 contrains one mole nitrogen atoms and 3 mole hydrogen atoms. Each nitrogen atom contains7 electrons and each hydrogen atom contains one electrons. No. of electons in 1 mole NH3 = 7 × 6.02 × 1023 + 3 × 6.02 × 1023

= 6.02 × 1024

5. A quantum is a packet of energy of definite magnitude, which may or may not be light energy. A photon is apacket of light energy having a difinite magnitude. Thus, a quantum may not be a photon but photon is aquantum.

6. The atomic number of the element is equal to the number of protons = 35Thus, the element is bromine (Br)

Mass number = number of protons + number of neutrons = 35 + 45 = 80Since the species contains one electrons more than the number of protons, it carries one unit negative charge.

The symbol of the species is Br8035

7. Isobar : N147 , Isotone : O16

8 , Isotope : C136

8. m = – 3, –2, –1, 0, +1, +2, +3.

9. 2 (It is an orbital)

10. n + for 3d = 5n + for 4s = 4so energy of 3d is higher.

11. The increasing order of frequency is :Radiation from FM radio < radiation from microwave oven < amber light from traffic signal < X rays < cosmicrays from outer space.


12. E1 for H atom = – I.E. = –2.18 × 10–18 JE1 for He+ = E1(H) × Z2 = –2.18 × 10–18 × 22 J

= – 8.72 × 10–18 J

The given process represents ionization of He+.

The energy required for ionization of He+ = )He(1E = 8.72 × 10–18 J.

13. = .E.Km2h

= 1327

34

10425.7106.62

106.6

= 6.67 × 10–15 m.

14. x × mV = 4

h

10 × 10–12 × 9.1 × 10–31 × V =

14.34106.6 34

V = 6 × 10–6 ms–1.

15. (a) For N shell principal quantum number, n = 4Number of sub-shells in an energy level = n Number of sub-shells in N shell = 4(b) For d subshell = 2Number of orbitals in a sub-shell = 2 + 1Number of orbitals in a d-sub shell = (2 × 2) + 1 = 5

16. NeNa 2210

2211

There is change in atomic number of one unit, hence, it can

(i) capture and electron NeeNa 2210

01

2211

(ii) emit a positron 01

2210

2211 NeNa

17. (i) 16 (ii) 7

18. Sphrical node = n – 1–(i) 2 (ii) 1

19. Cr+3 = 1s2 2s2 2p6 3s2 3p6 4s0 3d3 Ans. 3

20. m = )ms103)(m101(

smkg10626.6c

h189

1234

= 2.208 × 10–33 kg.

21. Three, principal quantum number, Azimuthal quantum number and Magnetic quantum number.

22. PbU 20782

23592 + 7 He4

2 + 4 e01 (4n + 3 series)


23. s- orbital : Shape spherical

Size of 2s is more than 1s and 2s contain 1 radial node while 1s does not.

24. Fe atom has 26 electrons. Fe2+ and Fe3+ ions are formed by removal of two and three electrons respectivelyfrom Fe atom and hence contain 24 and 23 electrons respectively. Their electronic configuration are :

Fe2+ : 1s2, 2s2, 2p6, 3s2, 3p6, 3d6

Fe3+ : 1s2, 2s2, 2p6, 3s2, 3p6, 3d5

Fe2+ has 4 unpaired electrons while Fe3+ has 5 unpaired electrons. Therefore, Fe3+ has more number ofunpaired electrons.

25. According to de-Broglie, the wavelength associated with a particle of mass m, moving with velocity v is givenby the relation,

= mvh

where h is Planck’s constant

21

mv2 = KE

m2v2 = 2m KEmv = mKE2

= mKE2h

26. The exact position and momentum of a fast moving particle cannot be calculated precisely at the samemoment of time. If x is the error in the measurement of position of the particle and if p is the error in themeasurement of momentum of the particle, then:

x . p 4h

or x . (mv) 4

h

where, x = uncertainty in positionp = uncertainty in momentumh = Plank’s constantm = mass of the particlev = uncertainty in velocity.


27. p - orbital : Shape dumb bellDumb bell shape consists of two lobes which are separated by a region of zero probability called node.

p - sub shell can accomodate maximum no. of six electrons.

d - orbital : shape double dumb bell

d - sub shell can accomodate maximum no. of 10 electrons.

28. The key to writting nuclear equations is to make sure that the number of nucleons is the same on both sidesof the equation and that the number of elementary and nuclear charges is the same.

(a) Cm24296 PuHe 238

9442 (b) Mg28

12 Ale 2813

01

(c) Xe11854 I118

5301 e

29. (i) 2s (ii) 4d (iii) 3p

30. Hund’s rule :No electron pairing takes place in the orbitals in a sub - shell until each orbital is occupied byone electron with parallel spin. Exactly half filled and fully filled orbitals make the atoms more stable, i.e., p3, p6,d5, d10, f7 and f14 configuration are most stable.


N = 1s2 2s2 2p3

31. Production of Anode rays (Discovery of Proton) :

Goldstein (1886) performed the experiment with a discharge tube filled with a perforated cathode and foundthat new type of rays came out through the hole in the cathode.

When this experiment is conducted, a faint red glow is observed on the wall behind the cathode. Since theserays originate from the anode, they are called anode rays.

32. Drawbacks of Rutherford’s model :1. This was not according to the classical theory of electromagnetism proposed by Maxwell. According

to this theory, every accelerated charged particle must emit radiations in the form of electromagneticwaves and lose its total energy.Since energy of electrons keep on decreasing, so radius of the circular orbits should also decreaseand ultimately the electron should fall in nucleus.

2. It could not explain the line spectrum of H-atom.

33. E = N h

= nh c

= )m101.337()ms100.3()Js10626.6()106.5(

9

183424

= 3.3 × 106 J s–1

34. (i) The phenomenon of splitting up of a heavy nucleus, on bombardment with slow speed neutrons, into twofragments of comparable mass, with the release of two or more fast moving neutrons and a large amount ofenergy, is known as nuclear fission.

UnU23692

10

23592

(ii) The phenomenon of joining up of two light nuclei into a heavier nucleus is called fusion, e.g.,

nHeHH 10

42

31

21 ; (H = –17.6 MeV)

(iii) The total energy given out during binding up of nucleons in nucleus is known as binding energy. Greater isthe binding energy, lesser will be the energy level of nucleus and thus, more will be its stability.


35. Assumptions of Bohr’s model :

There are certain orbits around the nucleus such that if electron will be revolving in these orbit, withoutemmitting any electromagnetic radiation. These are called stationary orbit.The neccessary centripetal force is produced by attraction forces of nucleus.

r

mv2 = 2

2

rZKe

Angular momentum of the electron in these stationary orbit is always an integral multiple of 2h

mvr = 2

nh

Electron can make jump from one stationary orbit to another stationary orbit by absorbing or emittinga photon of energy equal to difference in the energies of the stationary orbit.

hc

= E E = difference in the energy of orbit

Line Spectrum of Hydrogen :Line spectrum of hydrogen is observed due to excitation or de-excitation of electron from one stationary orbitto another stationary orbitLet electron makes transition from n2 to n1 (n2 > n1) in a H-like sample

photon

n1

n2eV– 13.6 Z2

n22

– 13.6 Z2

n12 eV

Energy of emitted photon = (E)n2 n1 = 22

2

nZ6.13

–

21

2

nZ6.13

= 13.6Z2

2

22

1 n1

n1

Wavelength of emitted photon

= 12 nn)E(

hc

=

2

22

1

2

n1

n1Z6.13

hc

1

=

2

22

1

2

n1

n1

hcz)6.13(

Wave number,1

= =

2

221

2

n1

n1RZ

R = Rydberg constant = 1.09678 × 107m–1 ;

R ~1.1 × 107 m–1 ; R = hceV6.13

; R ch = 13.6 eV


36. Fe+3 = 1s2 2s2 2p6 3s2 3p6 3d5

(i) 2p and 3s have n + = 3 so number of electrons = 8.(ii) It is IIIrd shell which contains three orbitals having m = 0 so number of electrons = 5.(iii) It is p sub shell so number of electron = 12. (iv) For M-shell n = 3 so number of electron = 13.

37. A (Zero), B(1st), C(2nd), D(Zero)A and D are isobtopes.A, B and C are isobars.

38. Cr = 1s2 2s2 2p6 3s2 3p6 4s1 3d5

(i) sub-shells = 1s, 2s, 2p, 3s, 3p, 4s and 3d so total subshell = 7(ii) orbitals = 1 + 1 + 3 + 1 + 3 + 1+ 5 = 15(iii) M-shell have n = 3 so elctron = 13 Ans. (i) 7 (ii) 15 (iii) 13

39. The total energy of an electron revolving in a particular orbit isT.E. = K.E. + P.E.

where :P.E. = Potential energy , K.E. = Kinetic energy , T.E. = Total energy

The K.E. of an electron = 21

mv2

and the P.E. of an electron = – r

KZe2

Hence, T.E. = 21

mv2 – r

KZe2

we know that,r

mv2 = 2

2

rKZe

or mv2 = r

KZe2

substituting the value of mv2 in the above equation :

T.E. =r2

KZe2–

rKZe2

= – r2

KZe2

So, T.E. = – r2

KZe2

substituting the value of ‘r’ in the equation of T.E.

Then T.E. = – 2

KZe2 x 22

22

hnmZe4

= – 22

2422

hnKmeZ2

Thus, the total energy of an electron in nth orbit is given by

T.E. = En = – 2

242

hkme2

2

2

nz

... (iv)

Putting the value of m,e,h and we get the expression of total energy

En = – 13.6 2

2

nZ

eV / atom n T.E. ; Z T.E.


EXERCISE - 1PART-I

A-1.

Particle Atomic No. Mass No. No. of electrons No. of protons No. of neutrons

Sodium atom 11 23 11 11 12

Aluminium ion 13 27 10 13 14

Chloride ion 17 35 18 17 18

Phosphorus atom 15 31 15 15 16

Cuprous ion 29 64 28 29 35

A-2. 1.8 × 10–43 m3 A-3. 2.7 × 10–14 A-4. (A) 6.5 × 10–15 m , (B) 2

2

vmKe188

B-1. 621.1 eV. B-2. 1.56 × 1016 B-3. 219.3 m, 4.56 × 10–3 m–1

B-4. 239.4 KJ/mol. B-5. 1.35 × 105 photons B-6. 200 watt.

B-7. No B-8. 1 × 1016 Hz B-9. (a) 4000 Å (b) 1.033 ev

C-1. n = 2 C-2. 7.27 × 105 m/s C-3. x = 2

C-4. 329

C-5. 4.36 × 106 m/s, 0.0145, 3.8 × 10–17 sec, 2.63 × 1016 revolution

C-6. ‘x’ C-7. A = 2, B = 4 C-8. n = 3

C-9. (a) Z = 3, (b) 108.8 eV, (c) 1.013 × 10–8 m , (d) 122.4 eV

C-10. (a) 15.6 V , (b) 233.9 nm, (c) 1.009 × 107 m–1

C-11. 54.4 eV D-1. 6561 Å, 4863 Å (Approx) D-2. n = 4 to n = 2

D-3. v = 7.3 × 1014 Hz, visible spectrum D-4. z = 2

D-5. 182.5 kJ D-6. 20 E-1. 4.71 Å

E-2. ve = 1836 vp E-3. 6320

E-4. 6.15 Å

E-5. V = m21

h

E-6. ~ 100 gm F-1. 2

F-2. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2. F-3. 3p, 5d, 4p, 2s, 4d

F-4. (a) 0, (b) 2h

, (c) h2

F-5. Impossible sets of quantum numbers are (i), (iii), and (vi)(i) = 0 m = 0 (m 1) (iii) n = 1 = 0 ( 2) (vi) s = + 1/2 or – 1/2 (s 0)

F-6. (i) + 5/2 or – 5/2, spin magnetic moment = 35 B.M. (ii) 0, 0

G-1. E = m × 931.478

m = 478.931E

= 478.93125.17

= 0.0185 amu

m = 3.07 × 10–26 g.


G-2. Mg2412 + n1

0 Na2411

+ H11

G-3. (a) N147 + n1

0 C146 + H1

1

(b) K3919 + n1

0 Cl3617 + He4

2

G-4. (a) Ca3820 : It has p

n =

2018

= 0.9, Which lies below the belt of stability and thus positron emitter..

Ca3820 K38

19 + e01

(b) Ar3518 : It has p

n =

1817

= 0.994, which lies below the belt of stability and thus, positron emitter

Ar3518 Cl35

17 + e01

If pn

< 1 and nuclear charge is high the nuclide may show K-electron capture.

(c) Ge8032 : It has p

n =

3248

= 1.5, which lies above the belt of stability and thus -emitter..

Ge8032 As80

33 + e01

(d) Ca4020 : It has both magic numbers p = 20, n = 20 and thus, stable.

G-5. 3n, 3n – 1 EPART-II

A-1. (A) A-2. (D) A-3. (A) A-4.* (A,B) A-5. (C)

B-1. (B) B-2. (A) B-3. (C) B-4. (A) B-5. (C)

B-6. (D) B-7. (D) C-1. (A) C-2. (B) C-3. (B)

C-4. (C) C-5. (A) C-6. (A) C-7. (C) C-8. (C)

C-9. (C) C-10. (B) C-11.* (A,B,D) D-1. (A) D-2. (B)

D-3. (D) D-4.* (A) D-5. (C) D-6. (C) D-7. (D)

D-8. (D) E-1. (D) E-2. (B) E-3. (C) E-4. (C)

E-5. (B) E-6. (A) E-7. (C) F-1. (A) F-2. (D)

F-3. (A) F-4. (B) F-5. (D) F-6. (A) F-7. (B)

F-8. (B) F-9. (D) F-10. (C) F-11. (C) F-12.* (A,B,C)

G-1 (C) G-2 (A) G-3 (D) G-4 (C) G-5 (C)

G-6 (C) G-7 (C) G-8 (A) G-9 (A) G-10 (D)

H-1* (A,B,D) H-2 (A) H-3* (A,B,D) H-4 (C) H-5 (C)

H-6 (A) H-7 (B) H-8 (B) H-9 (D)

PART-III1. (D) 2. (A) 3. (A) 4. (A) 5. (C)


EXERCISE - 2PART-I

1. 3.32 × 106 J

2. n = 3.75 × 1019

3. 18 × 1022

4. From Heisenberg’s uncertainty principle

x × p = 4h

or x × (mv) = 4h

v = xm4h

The value of x will be of the order of size of nucleus i.e. 10–15 m.m = mass of electron = 9.109 × 10–31 kg. Putting these values, we get

v = 1531

34

1010109.914.3410625.6

5.785 × 1010 m/s.

However, the maximum value of v can be 3 × 108 m/s (speed of light)Thus, the obtained value of v is not practically possible.Hence, it can be said that the moving electron cannot exist within the nucleus.

5. 3 2 = 6565 Å, 2 1 = 1216Å, 3 1 = 1026 Å

6. 12

7. (a) n = 2 (b) 14.4 eV (c) 13.5 eV, 0.699 eV

8. En – En –1 = h, where is the frequency of radiation emitted by H-atom due to transition of electron from n to(n – 1)

=

223

2242

n1

)1n(1

hzkme2

=

223

2242

)1n(n1n2

hzkme2

=

23

2242

)1n(nn/12

hzkme2

For very large values of n, 1/n 0 & (n – 1) n

= 33

2242

hnzkme4

.... (i)

The frequency of revolution of electron in nth orbit, fn is given by :

fn = n

nr2

v =

zkme4hn2

nhkze2

22

22

2

= 33

2242

hnzkme4

= 33

2242

hnzkme4

..... (ii)

From (i) & (ii) for very large values of n, the frequency of radiation emitted by H-atom due to transition ofelectron from n to (n – 1) is equal to frequency of revolution of electron in its orbit.

9. x = 8 Ans.

10. n2 = 12


PART-II1. (A) 2. (C) 3. (C) 4. (D) 5. (D)

6. (D) 7. (D) 8. (B) 9. (C) 10. (B)

11. (A) 12. (A) 13. (B) 14. (D) 15. (C)

16. (A) 17. (B) 18. (A) 19. (D) 20. (A)

21. (A) 22. (C) 23. (B) 24. (A) 25. (D)

26. (B) 27. (A) 28. (C) 29. (C) 30. (A)

31. (A,C) 32. (B,D) 33. (A,B,C) 34. (A,C) 35. (A,B,C)

PART-III1. (i – e) ; (ii – d) ; (iii – f) ; (iv – b) ; (v – a) ; (vi – c) 2. (i – f) ; (ii – d) ; (iii – a) ; (iv – e) ; (v – b) ; (vi – c)

3. (i – b) , (ii – a), (iii – b, c) , (iv – c, d). 4. (i – c) , (ii – d), (iii – a), (iv – b).

PART-IV1. (B) 2. (C) 3. (A) 4. (A) 5. (D)

6. (C) 7. (D) 8. (D) 9. (C) 10. (C)

11. (B) 12. (B) 13. (B) 14. (B) 15. (A)

EXERCISE - 3PART-I

1. 98.17 kJ 2. (D)

3. (a) r = 2a0

(b) = 6.626 × 10–25 Å (c) 82Y206 ; (Atomic no. 82, Mass no. 206)

4. (a) 2.18 106 m/s, 3.32 × 10–10 m ; (b) 2 .

2h

5. [A — r] ; [B – q] ; [C – p] ; [D – s].

6. (B) 7. (C) 8. (B) 9. 4 10. 9

11. (C) 12. (A)

13. 8PART-II

1. (A) 2. (C) 3. (A) 4. (A) 5. (B)

6. (D) 7. (B) 8. (A) 9. (C) 10. (B)

11. (A) 12. (A) 13. (C) 14. (D) 15. (A)

16. (A) 17. (D) 18. (C) 19. (D) 20. (B)

21. (C) 22. (A) 23. (B)

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Atomic Structure Exercises by Resonance