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www.chemistryonline.in
CHAPTER 2: Structure of Atom
(CLASS: XI)
www.chemistryonline.in P – 1
NCERT Solved Examples Problem 2.1
Calculate the number of protons, neutrons and electrons in 8035Br .
Solution
In this case, 8035Br , Z = 35, A = 80, species is neutral
Number of protons = number of electrons = Z = 35
Number of neutrons = 80 – 35 = 45
Problem 2.2
The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper
symbol to the species.
Solution
The atomic number is equal to number of protons = 16. The element is sulphur (S).
Atomic mass number = number of protons + number of neutrons = 16 + 16 = 32
Species is not neutral as the number of protons is not equal to electrons. It is anion (negatively charged) with charge
equal to excess electrons = 18 – 16 = 2.
Symbol is 32 216 S .
Problem 2.3
The Vividh Bharati station of All India Radio, Delhi, broadcasts on a frequency of 1,368 kHz (kilo hertz). Calculate the
wavelength of the electromagnetic radiation emitted by transmitter. Which part of the electromagnetic spectrum does it
belong to?
Solution
The wavelength, , is equal to c/ , where c is the speed of electromagnetic radiation in vacuum and is the
frequency. Substituting the given values, we have
c 8 13.00 10 m s
1368 kHz
8 1
3 1
3.00 10 m s
1368 10 s= 219.3 m
This is a characteristic radiowave wavelength.
Problem 2.4
The wavelength range of the visible spectrum extends from violet (400 nm) to red (750 nm). Express these
wavelengths in frequencies (Hz). (1 nm = 10–9
m)
Solution
Using equation 2.5, frequency of violet light
8 1
9
c 3.00 10 m s
400 10 m= 7.50 10
14 Hz
Frequency of red light
8 1
9
c 3.00 10 ms
750 10 m = 4.00 10
14 Hz
The range of visible spectrum is from
4.0 1014
to 7.5 1014
Hz in terms of frequency units.
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CHAPTER 2: Structure of Atom
(CLASS: XI)
www.chemistryonline.in P – 2
Problem 2.5
Calculate (a) wavenumber and (b) frequency of yellow radiation having wavelength 5800 Å.
Solution
(a) Calculation of wavenumber ( )
= 5800 Å = 5800 10–8
cm = 5800 10–10
m
10
1 1
5800 10 m= 1.724 10
6 m
–1= 1.724 10
4 cm
–1
(b) Calculation of the frequency ( )
8 1
10
c 3 10 m s
5800 10 m = 5.172 10
14 s
–1
Problem 2.6
Calculate energy of one mole of photons of radiation whose frequency is 5 1014
Hz.
Solution
Energy (E) of one photon is given by the expression
E = h
h = 6.626 10–34
J s; = 5 1014
s–1
(given)
E = (6.626 10–34
J s) (5 1014
s–1
) = 3.313 10–19
J
Energy of one mole of photons = (3.313 10–19
J) (6.022 1023
mol–1
) = 199.51 kJ mol–1
Problem 2.7
A 100 watt bulb emits monochromatic light of wavelength 400 nm. Calculate the number of photons emitted per
second by the bulb.
Solution
Power of the bulb = 100 watt = 100 J s–1
Energy of one photon E = h = hc/ 34 8 1
9
6.626 10 Js 3 10 m s
400 10 m = 4.969 10
−19 J
Number of photons emitted= 1
19
100 J s
4.969 10 J = 2.012 10
20 s
–1
Problem 2.8
When electromagnetic radiation of wavelength 300 nm falls on the surface of sodium, electrons are emitted with a
kinetic energy of 1.68 105 J mol
–1. What is the minimum energy needed to remove an electron from sodium? What is
the maximum wavelength that will cause a photoelectron to be emitted ?
Solution
The energy (E) of a 300 nm photon is given by
h = hc/34 8 1
9
6.626 10 J s 3.0 10 m s
300 10 m= 6.626 10
–19 J
The energy of one mole of photons
= 6.626 10–19
J 6.022 1023
mol–1
= 3.99 105 J mol
–1
The minimum energy needed to remove one mole of electrons from sodium
= (3.99 –1.68) 105 J mol
–1 = 2.31 10
5 J mol
–1
The minimum energy for one electron 5 1
23 1
2.31 10 J mol
6.022 10 electrons mol = 3.84 10
−19 J
This corresponds to the wavelength
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CHAPTER 2: Structure of Atom
(CLASS: XI)
www.chemistryonline.in P – 3
ch
E
34 8 1
19
6.626 10 J s 3.0 10 m s
3.84 10 J = 517 nm
(This corresponds to green light)
Problem 2.9
The threshold frequency 0 for a metal is 7.0 1014
s–1
. Calculate the kinetic energy of an electron emitted when
radiation of frequency = 1.0 1015
s–1
hits the metal.
Solution
According to Einstein’s equation
Kinetic energy = ½ mev2 = h( – 0)
= (6.626 10–34
J s) (1.0 1015
s–1
– 7.0 1014
s–1
)
= (6.626 10–34
J s) (10.0 1014
s–1
– 7.0 1014
s–1
)
= (6.626 10–34
J s) (3.0 1014
s–1
)
= 1.988 10–19
J
Problem 2.10
What are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n = 2 state in the
hydrogen atom?
Solution
Since ni = 5 and nf = 2, this transition gives rise to a spectral line in the visible region of the Balmer series. From
equation (2.17)
18
2 2
1 12.18 10 J
5 2E = – 4.58 10
–19 J
It is an emission energy
The frequency of the photon (taking energy in terms of magnitude) is given by
E
h
19
34
4.58 10 J
6.626 10 Js= 6.91 10
14 Hz
8 1
14
c 3.0 10 m s
6.91 10 Hz = 434 nm
Problem 2.11
Calculate the energy associated with the first orbit of He+. What is the radius of this orbit?
Solution 18 2
n 2
(2.18 10 J)ZE
n atom
–1
For He+, n = 1, Z = 2
18 2
1 2
(2.18 10 J) (2)E
1 = – 8.72 10
–18 J
The radius of the orbit is given by 2
n
(0.0529 nm)r
n
Z
Since n = 1, and Z = 2 2
n
(0.0529 nm) lr
2 = 0.02645 nm
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CHAPTER 2: Structure of Atom
(CLASS: XI)
www.chemistryonline.in P – 4
Problem 2.12
What will be the wavelength of a ball of mass 0.1 kg moving with a velocity of 10 m s–1
?
Solution
According to de Broglie equation (2.22) 34
1
h (6.626 10 Js)
mv (0.1kg) (10 m s )= 6.626 10
–34 m (J = kg m
2 s
–2)
Problem 2.13
The mass of an electron is 9.1 10–31
kg. If its K.E. is 3.0 10–25
J, calculate its wavelength.
Solution
Since K. E. = ½ mv2
1/21/2 25 2 2
31
2 K.E. 2 3.0 10 kg m sv
m 9.1 10 kg= 812 m s
–1
34
31 1
h 6.626 10 Js
mv (9.1 10 kg) (812 ms )= 8967 10
–10 m = 896.7 nm
Problem 2.14
Calculate the mass of a photon with wavelength 3.6 Å.
Solution
= 3.6 Å = 3.6 10−10
m
Velocity of photon = velocity of light 34
10 8 1
h 6.626 10 Jsm
(3.6 10 m) (3 10 m s )= 6.135 10
–29 kg
Problem 2.15
A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Å. What is
the uncertainty involved in the measurement of its velocity?
Solution
hx p
4 or
hxm v
4 or
hv
4 xm
34
10 31
6.626 10 Jsv
4 3.14 0.1 10 m 9.11 10 kg= 0.579 10
7 m s
–1 (1J = 1 kg m
2 s
–2) = 5.79 10
6 m s
–1
Problem 2.16
A golf ball has a mass of 40g, and a speed of 45 m/s. If the speed can be measured within accuracy of 2%, calculate
the uncertainty in the position.
Solution
The uncertainty in the speed is 2%, i.e.,
245
100 = 0.9 m s
–1
hx
4 m v
34
3 1 1
6.626 10 Js
4 3.14 40 g 10 kg g (0.9 ms )= 1.46 10
–33 m
This is nearly ~ 1018
times smaller than the diameter of a typical atomic nucleus. As mentioned earlier for large
particles, the uncertainty principle sets no meaningful limit to the precision of measurements.
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CHAPTER 2: Structure of Atom
(CLASS: XI)
www.chemistryonline.in P – 5
Problem 2.17
What is the total number of orbitals associated with the principal quantum number n = 3 ?
Solution
For n = 3, the possible values of l are 0, 1 and 2. Thus there is one 3s orbital (n = 3, l = 0 and ml = 0); there are three
3p orbitals (n = 3, l = 1 and ml = –1, 0, +1); there are five 3d orbitals (n = 3, l = 2 and ml = –2, –1, 0, +1, +2).
Therefore, the total number of orbitals is 1 + 3 + 5 = 9
The same value can also be obtained by using the relation; number of orbitals = n2, i.e. 3
2 = 9.
Problem 2.18
Using s, p, d, f notations, describe the orbital with the following quantum numbers
(a) n = 2, l = 1,
(b) n = 4, l = 0,
(c) n = 5, l = 3,
(d) n = 3, l = 2
Solution
n l orbital
(a) 2 1 2p
(b) 4 0 4s
(c) 5 3 5f
(d) 3 2 3d
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CHAPTER 2: Structure of Atom
(CLASS: XI)
www.chemistryonline.in P – 6
NCERT Exercise
2.1 (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Ans: (i) 9.1 10–31
kg = 9.1 10–28
kg = Mass of 1 electron
9.1 10–28
g = 1 electron
1 g =28101.9
1 electrons = 1.1 10
–27 electrons
(ii) Mass of 1 electron = 9.1 10–31
kg
Mass of 1 mole electrons = 6.023 1023
9.1 10–31
kg
= 5.48 10–7
kg
Charge on 1 electron = – 1.6 10–19
C
Charge on 1 mole electrons = 6.023 1023
(– 1.6 10–19
) C
= – 96368 C
2.2 (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14
C.
(Assume that mass of a neutron = 1.675 10–27
kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.
Will the answer change if the temperature and pressure are changed?
Ans: (i) Total electrons in 1 molecule of CH4 = 10 electrons
Electrons in 1 mole of CH4 = 10 mole electrons
= 10 6.023 1023
electrons
= 6.023 1024
electrons
(ii) (a) 1 molecule of 14
C has = (14 – 6) = 8 neutrons
1 mole of 14
C has = 8 mol neutrons
Now, 1 mol of 14
C = 14 g 14
C
14 g of 14
C = 8 mol neutrons
1 g of 14
C = 8
14mol neutrons
7 mg of 14C = 0.007 g of 14
C = 8
0.00714
mol neutrons
= 0.004 mol neutrons
= 0.004 6.023 1023
neutrons
= 2.41 1021
neutrons
(b) Mass of 1 neutron = 1.67 10–27
kg
Total mass of neutrons = 1.67 10–27
2.41 1021
kg
= 4.02 10–6
kg
2.3 How many neutrons and protons are there in the following nuclei?
13 16 24 56 88
6 8 12 26 38C, O, Mg, Fe, Sr
Ans: Neutrons Protons 136 C 7 6
168 O 8 8
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CHAPTER 2: Structure of Atom
(CLASS: XI)
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2412 Mg 12 12
5626Fe 30 26
8838Sr 50 38
2.4 Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17 , A = 35.
(ii) Z = 92 , A = 233.
(iii) Z = 4 , A = 9.
Ans: (i) 3517 Cl (ii) 233
92 U (iii) 94Be
2.5 Yellow light emitted from a sodium lamp has a wavelength ( ) of 580 nm. Calculate the frequency ( )
and wavenumber ( ) of the yellow light.
Ans: = 580 nm = 580 10–9
m
v = c or 9
8
10580
103c = 5.17 10
14 Hz
Wave number = 910580
11= 1.72 10
6 m
–1
2.6 Find energy of each of the photons which
(i) correspond to light of frequency 3 1015
Hz.
(ii) have wavelength of 0.50 Å.
Ans: (i) E = hv = 6.6 10–34
3 1015
= 1.98 10–18
J
(ii) 10
834
1050.0
103106.6hc E = 3.96 10
–15 J
2.7 Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 10–10
s.
Ans: Frequency, v =10102
1
period time
1 = 5 10
9 Hz
v = c or =9
8
105
103
v
c = 0.06 m
Wave number = 06.0
11= 16.67 m
–1
2.8 What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?
Ans: nhc
E
or 34 8
12
n 6.6 10 3 101
4000 10
or n = 2.02 1016
photons
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CHAPTER 2: Structure of Atom
(CLASS: XI)
www.chemistryonline.in P – 8
2.9 A photon of wavelength 4 10–7
m strikes on metal surface, the work function of the metal being 2.13
eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the
velocity of the photoelectron (1 eV = 1.6020 10–19
J).
Ans: = 4 10–7
m
Work function = 2.13 eV
(i) Energy of photon,
E = hv = 34 8
7
hc 6.6 10 3 10
4 10
J
= 4.95 10–19
J
19
19
4.95 10
1.6 10
eV
= 3.1 eV
(ii) Kinetic energy = Energy of photon – Threshold energy
= (3.1 – 2.13) eV = 0.97 eV
(iii) K.E. = 1
2mv
2, where m = 9.1 10
–31 kg (mass of electron)
K.E. = 0.97 eV = 0.97 1.6 10–19
J = 1.55 10–19
J
1.55 10–19
= 1
2(9.1 10
–31) (v
2)
or v2 =
19
31
2 1.55 10
9.1 10
= 3.4 10
11
or v = 113.4 10 = 5.83 105 m/s
2.10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom.
Calculate the ionisation energy of sodium in kJ mol–1
.
Ans: Wavelength of radiation, = 242 nm = 242 10–9
m
K.E. = 0
hv = hv0 = ionization energy
Ionisation energy= hv = 34 8
9
hc 6.6 10 3 10
242 10
= 8.2 10
–19 J
Ionisation energy in kJ/mol = 198.2 10
1000
6.023 10
23
= 493.88 kJ/mol
2.11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 m. Calculate the rate of
emission of quanta per second.
Ans: Power = 25 watt
Power = Energy
Time (Here, time = 1 s)
25 = E
1 or E = 25 J
Now, E = nhv = nhc
Given, = 0.57 m = 0.57 10–6
m ; n = ?
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CHAPTER 2: Structure of Atom
(CLASS: XI)
www.chemistryonline.in P – 9
25 = 34 8
6
n 6.6 10 3 10
0.57 10
6
34 8
25 0.57 10n
6.6 10 3 10
= 7.2 10
19 photons (quanta) per second
2.12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of
wavelength 6800 Å. Calculate threshold frequency ( 0) and work function (W0) of the metal.
Ans: Velocity, v = zero
Wavelength, = 6800 Å = 6800 10–10
m
Frequency, 8
10
c 3 10v
6800 10
= 4.4 1014
Hz
Now, hv = hv0 + 1
2mv
2
As v = 0 : hv = hv0
or v = v0 = 4.4 1014
Hz
Threshold frequency, v0 = 4.4 1014
Hz
Work function = hv0 = 6.6 10–34
4.4 1014
= 2.904 10–19
J
2.13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition
from an energy level with n = 4 to an energy level with n = 2?
Ans: 2 21 2
1 1v R
n n
z2 = 109677
2 2
1 1
2 4
(1)
2 cm
–1
= 109677 1 1
4 16
cm
–1
= 109677 3
16cm
–1
Now, 1
v
or = 1 16
v 3 109677
cm
= 4.86 10–5
cm
= 486 10–7
cm
= 486 nm
2.14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your
answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1
orbit).
Ans: Energy required = 2 2
2 2
13.6 z 13.6 (1)
n (5)
= 0.544 eV
Ionisation energy of H–atom = 13.6 eV
Comparison : 13.6
0.544 = 25 times
Ionisation energy of H–atom is 25 times the energy required to remove electron from 5th orbit.
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CHAPTER 2: Structure of Atom
(CLASS: XI)
www.chemistryonline.in P – 10
2.15 What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops
to the ground state?
Ans: Maximum number of emission lines 2 1 2 1(n n )(n n 1) (6 1)(6 1 1)
2 2
= 15 lines
2.16 (i) The energy associated with the first orbit in the hydrogen atom is –2.18 10–18
J atom–1
. What
is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Ans: (i) En = 18 2
2
2.18 10 z
n
J atom
–1
E5 = 18 2
2
2.18 10 (1)
(5)
= – 8.72 10
–20 J atom
–1
(ii) 2
n
0.529 nr Å
z
20.529 (5)
1
= 13.225 Å
2.17 Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic
hydrogen.
Ans: Longest wavelength transition corresponds to first line of Balmer series. (n1 = 2 to n2 = 3)
2
2 21 2
1 1v R (z)
n n
= 109677 2 2
1 1
2 3
= 109677
5
36
cm–1
or v 15232.9 cm–1
2.18 What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr
orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns
to the ground state? The ground state electron energy is –2.18 10–11
ergs.
Ans: Energy required to shift electron from 1st Bohr orbit to fifth Bohr orbit, E = E5 – E1
E = 18 18
2 2
2.18 10 2.18 10
(5) (1)
= 2.18 10–18
1
125
= 2.18 10–18
24
25
= 2.09 10–18
J
or 2.09 10–11
erg (1 J = 107 erg)
Now, for calculating wavelength,
hc
E
or hc
E
34 8
18
6.6 10 3 10
2.09 10
= 9.47 10
–8 m
= 947 10–10
m
= 947 Å
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CHAPTER 2: Structure of Atom
(CLASS: XI)
www.chemistryonline.in P – 11
2.19 The electron energy in hydrogen atom is given by En = (–2.18 10–18
)/n2 J. Calculate the energy
required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of
light in cm that can be used to cause this transition?
Ans: Energy required to remove electron from n = 2 is 18 2
2
2.18 10 z
n
18
2
2.18 10
(2)
= 0.59 10
–18 J atom
–1
Now, E = hc
or 34 8
19
hc 6.6 10 3 10
E 5.9 10
= 3.35 10
–7 m
= 3.35 10–5
cm
= 3350 10–8
cm
= 3350 Å
2.20 Calculate the wavelength of an electron moving with a velocity of 2.05 107 m s
–1.
Ans: = ? ; Mass of electron, m = 9.1 10–31
kg ;
v = 2.05 107 m/s
34
31 7
h 6.6 10
mv 9.1 10 2.05 10
= 3.53 10
–11 m
2.21 The mass of an electron is 9.1 10–31
kg. If its K.E. is 3.0 10–25
J, calculate its wavelength.
Ans: h
mv
Now, K.E. = 21
mv2
or 2 2KE
vm
or 2KE
vm
Now, 22KE 2m KE
mv m 2mKEm m
h h
mv 2mKE
m = 9.1 10–31
kg ; KE = 3.0 10–25
J
34
31 25
6.6 10
2 9.1 10 3.0 10
= 8.93 10–7
m
= 8930 10–10
m
= 8930 Å
2.22 Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na+, K
+, Mg
2+, Ca
2+, S
2–, Ar.
Ans: Na+, Mg
2+ : iso–electronic having 10 electrons each
K+, Ca
2+, S
2–, Ar : iso–electronic having 18 electrons each
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CHAPTER 2: Structure of Atom
(CLASS: XI)
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2.23 (i) Write the electronic configurations of the following ions: (a) H– (b) Na
+ (c) O
2– (d) F
–
(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1
(b) 2p3 and (c) 3p
5 ?
(iii) Which atoms are indicated by the following configurations ?
(a) [He] 2s1 (b) [Ne] 3s
2 3p
3 (c) [Ar] 4s
2 3d
1.
Ans: (i) (a) H– : 1s
2 (b) Na
+ : 1s
22s
22p
6
(c) O2–
: 1s22s
22p
6 (d) F
– : 1s
22s
22p
6
(ii) (a) Z = 11 (b) Z = 7 (c) Z = 17
(iii) (a) Li (b) P (c) Sc
2.24 What is the lowest value of n that allows g orbitals to exist?
Ans: n = 5
2.25 An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
Ans: n = 3, l = 2, ml = – 2, – 1, 0, +1, +2
2.26 An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and
(ii) the electronic configuration of the element.
Ans: (i) Number of protons = No. of electrons = 29
(ii) Electronic configuration: 1s22s
22p
63s
23p
63d
104s
1
2.27 Give the number of electrons in the species 2 2H , H and
2O
Ans: H2+ : 1 electron, H2 : 2 electrons, O2
+ : 15 electrons
2.28 (i) An atomic orbital has n = 3. What are the possible values of l and ml ?
(ii) List the quantum numbers (ml and l) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible?
1p, 2s, 2p and 3f
Ans: (i) n = 3, l = 0 ml = 0
l = 1 ml = – 1, 0, +1
l = 2 ml = – 2, – 1, 0, +1, +2
(ii) n = 3, l = 2, ml = – 2, – 1, 0, +1, +2
(iii) 2s, 2p
2.29 Using s, p, d notations, describe the orbital with the following quantum numbers.
(a) n = 1, l = 0; (b) n = 3; l = 1
(c) n = 4; l = 2; (d) n = 4; l = 3.
Ans: (a) 1s (b) 3p (c) 4d (d) 4f
2.30 Explain, giving reasons, which of the following sets of quantum numbers are not possible.
(a) n = 0, l = 0, ml = 0, ms = + ½
(b) n = 1, l = 0, ml = 0, ms = – ½
(c) n = 1, l = 1, ml = 0, ms = + ½
(d) n = 2, l = 1, ml = 0, ms = – ½
(e) n = 3, l = 3, ml = –3, ms = + ½
(f) n = 3, l = 1, ml = 0, ms = + ½
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CHAPTER 2: Structure of Atom
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Ans: (a) Not possible because ‘n’ can not be zero.
(b) Possible
(c) Not possible because ‘n’ and ‘l’ can not be zero.
(d) Possible
(e) Not possible because ‘n’ and ‘l’ can not be equal
(f) Possible
2.31 How many electrons in an atom may have the following quantum numbers?
(a) n = 4, ms = – ½ (b) n = 3, l = 0
Ans: (a) 16 electrons (b) 2 electrons
2.32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de
Broglie wavelength associated with the electron revolving around the orbit.
Ans: According to Bohr’s model
nhmvr
2
or nh
mv2 r
…(1)
According to de–Broglie’s principle
h
mv …(2)
Putting the value of ‘mv’ from (1) in (2)
h
2 rnh
or 2 r
n
or 2 r = n
Hence proved
2.33 What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition
n = 4 to n = 2 of He+ spectrum?
Ans: Let transition in hydrogen spectrum be n1 to n2
H 2 21 2
1 1v R
n n
z2 = R
2 21 2
1 1
n n
…(1)
Hl 2 2
1 1v R
2 4
(2)
2 = R
1 1
4 16
4 = R
4 4
4 16
or Hl
1 1v R
1 4
…(2)
Given that (1) = (2)
2 21 2
1 1 1 1R R
1 4n n
or 2 21 2
1 1 1 1
1 4n n
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or 21n 1 or n1 = 1
22n 4 or n2 = 2
2.34 Calculate the energy required for the process
He+ (g) He
2+ (g) + e
–
The ionization energy for the H atom in the ground state is 2.18 10–18
J atom–1
Ans: He+ He
2+ + e
–
Ionization energy of helium = 18 2
2
2.18 10 Z
n
18 2
2
2.18 10 (2)
(1)
= 8.72 10
–18 J
2.35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be
placed side by side in a straight line across length of scale of length 20 cm long.
Ans: Number of carbon atoms = 20 cm
0.15 nm 7
20 cm
0.15 10 cm
= 1.33 10
9
2.36 2 108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length
of this arrangement is 2.4 cm.
Ans: Diameter of 1 carbon atom = 8
2.4
2 10 = 1.2 10
–8 cm
Radius of 1 carbon atom = 81.2 10
2
= 6 10
–9 cm = 0.06 nm
2.37 The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms
present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.
Ans: Diameter = 2.6 Å
(a) Radius = 2.6
2 = 1.3 Å = 1.3 10
–10 m = 130 10
–12 m = 130 pm
(b) No. of atoms in a length of 1.6 cm 8
1.6 cm 1.6 cm
2.6 Å 2.6 10 cm
= 6.15 10
7 atoms
2.38 A certain particle carries 2.5 10–16
C of static electric charge. Calculate the number of electrons
present in it.
Ans: Charge on 1 electron = 1.6 10–19
C
or 1.6 10–19
C = 1 electron
1C = 19
1
1.6 10 electrons
2.5 10–16
C = 19
1
1.6 10 2.5 10
–16 electrons = 1563 electrons
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2.39 In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If
the static electric charge on the oil drop is –1.282 10–18
C, calculate the number of electrons present
on it.
Ans: Charge on 1 electron = 1.6 10–19
C
1.6 10–19
C = 1 electron
1 C = 19
1
1.6 10 electrons
1.282 10–18
C = 19
1
1.6 10 1.282 10
–18 electrons = 8 electrons
2.40 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been
used to be bombarded by the -particles. If the thin foil of light atoms like aluminium etc. is used,
what difference would be observed from the above results?
Ans: Smaller number of –particles will be deflected, because of lesser number of protons in lighter nucleus.
2.41 Symbols 79
35Br and 79
Br can be written, whereas symbols 79
35Br and 35
Br are not acceptable. Answer
briefly.
Ans: Try yourself.
2.42 An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the
atomic symbol.
Ans: Let number of protons be x
Number of neutrons = x + 31.7
100x
Mass number = No. of protons + No. of neutrons = 81
or x + x + 31.7x
100 = 81
31.7x2x 81
100
231.7x = 8100
or 8100
x 35231.7
No. of protons = Atomic number = 35
Element = Br
Symbol : 8135Br
2.43 An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more
neutrons than the electrons, find the symbol of the ion.
Ans: Mass no. = 37
Let no. of electrons be x
No. of neutrons = x + 11.1
100x
No. of protons = x – 1
Mass no. = No. of protons + No. of neutrons = 37
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or x – 1 + x + 11.1
100x = 37
211.1x – 100 = 3700
211.1x = 3600
x = 3600
211.1 = 17
No. of protons = x – 1 = 17 – 1 = 16
2.44 An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than
electrons. Assign the symbol to this ion.
Ans: Let no. of electrons be x
No. of neutrons = x + 30.4
x100
No. of protons = x + 3
Mass No. = 56 = No. of protons + No. of neutrons
x + 3 + x + 30.4
x 56100
230.4x + 300 = 5600
230.4x = 5300
x = 5300
230.4 = 23
No. of protons = x – 3 = 23 + 3 = 26
Symbol : 56 326Fe
2.45 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) cosmic rays from
outer space and (e) X-rays.
Ans: Radiation from FM radio < radiation from microwave oven < amber light from traffic signal < X–rays < cosmic
rays
2.46 Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is
5.6 1024
, calculate the power of this laser.
Ans: = 337.1 nm = 337.1 10–9
m
34 8
9
hc 6.6 10 3 10E
337.1 10
= 5.87 10
–19 J
Power = Energy
Time
(Here time is not given, so we assume t = 1 s)
Power = 195.87 10
1
= 5.87 10
–19 watt
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CHAPTER 2: Structure of Atom
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2.47 Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the
frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d)
number of quanta present if it produces 2 J of energy.
Ans: = 616 nm = 616 10–9
m
(a) Frequency of emission, 8
9
C 3 10v
616 10
= 4.87 1014
Hz
(b) Distance = velocity time
velocity of light = 3 108 m/s; time = 30 s
Distance = 3 108 30 = 9 10
9 m
(c) Energy of quantum, 34 8
9
hc 6.6 10 3 10E
616 10
J = 3.21 10
–19 J
(d) nhc
E
or 9
34 8
E 2 616 10n
hc 6.6 10 3 10
= 6.2 10
18 photons
2.48 In astronomical observations, signals observed from the distant stars are generally weak. If the
photon detector receives a total of 3.15 10–18
J from the radiations of 600 nm, calculate the number
of photons received by the detector.
Ans: nhc
E
or 3.15 10–18
= 34 8
9
n 6.6 10 3 10
600 10
or n = (3.15 10–18
600 10–9
)/6.6 10–34
3 108 = 9.54 10 photons
2.49 Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source
of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the
number of photons emitted during the pulse source is 2.5 1015
, calculate the energy of the source.
Ans: 9
1 1 1v
time period 2ns 2 10
= 5 10
8 Hz
n = 2.5 1015
Energy, E = nhv = 2.5 1015
6.6 10–34
5 108 = 8.25 10
–10 J
2.50 The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the
frequency of each transition and energy difference between two excited states.
Ans: 8
589 9
C 3 10v
589 10
= 5.093 1014
Hz
8
589.6 9
C 3 10v
589.6 10
= 5.088 1014
Hz
Energy difference = hv589.6 – hv589 = h (v589.6 – v589)
= 6.6 10–34
(5.093 1014
– 5.088 1014
)
= 6.6 10–34
1014
(5.093 – 5.088)
= 3.3 10–22
J
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2.51 The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the
threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm,
calculate the kinetic energy and the velocity of the ejected photoelectron.
Ans: Work function = 1.9 eV = 1.9 1.6 10–19
J = 3.04 10–19
J
(a) Threshold wavelength : 0
hcE
or 3.04 10–19
= 34 8
0
6.6 10 3 10
or 34 8
0 19
6.6 10 3 10
3.04 10
= 6.51 10
–7 m = 651 nm
(b) Threshold frequency
8
0 70
c 3 10v
6.51 10
= 4.61 1014
Hz
Wavelength of incident radiation = 500 nm = 500 10–9
m
Frequency of incident radiation, 8
19
c 3 10v
500 10
= 6 1014
Hz
Now, hv = hv0 + K.E.
K.E. = hv – hv0 = h (v – v0)
= 6.6 10–34
(6 1014
– 4.61 1014
)
= 6.6 10–34
1014
(6 – 4.61)
= 6.6 10–20
1.39 = 9.2 10–20
J
2.53 The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be
stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work
function for silver metal.
Ans: K.E. = eV = 1.6 10–19
0.35 = 0.56 10–19
J
= 256.7 nm = 256.7 10–9
m
8
9
c 3 10
256.7 10
= 1.17 10
15 Hz
hv = hv0 + K.E. or hv0 = hv – K.E.= (6.6 10–34
1.17 1015
) – (0.56 10–19
)
= (7.72 10–19
) – (0.56 10–19
)
= 10–19
(7.72 – 0.56)
= 7.16 10–19
J
19
19
7.16 10
1.6 10
= 4.3 eV
2.54 If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is
ejected out with a velocity of 1.5 107 m s
–1, calculate the energy with which it is bound to the
nucleus.
Ans: = 150 pm = 150 10–12
m
v = 1.5 107 m/s
Binding energy or work function, hv0 = ?
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CHAPTER 2: Structure of Atom
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hv = hv0 + 1
2mv
2
or hv0 = hv – 1
2mv
2 =
hc 1
2
mv
2
or hv0 = 34 8
12
6.6 10 3 10
150 10
31 7 219.1 10 (1.5 10 )
2
= [1.32 10–15
] – [0.1 10–15
] = 1.22 10–15
J
2.55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be
represented as = 3.29 1015
(Hz) [1/32 – 1/n
2]
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Ans: = 1285 nm = 1285 10–9
m,
v = 8
9
c 3 10
1285 10
= 2.33 10
14 Hz
2.33 1014
= 3.29 105
2
1 1
9 n
or 14
2 15
1 1 2.33 10
9 n 3.29 10
= 0.0708
2
1 1
9n – 0.0708 = 0.11 – 0.0708 = 0.0392
or n2 =
1
0.0392 = 25.57
or n = (25.57)1/2
5
2.56 Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm
and ends at 211.6 pm. Name the series to which this transition belongs and the region of the
spectrum.
Ans: r1 = 1.3225 nm = 1322.5 pm
r = 2 252.9 n 52.9 n
z 1
= 1322.5
or n2 =
1322.5
52.9 = 25 or n = 5
r2 = 211.6 pm
52.9 n2 = 211.6
n2 =
211.6
52.9 = 4 or n = 2
2
2 21 2
1 1v R Z
n n
= 109677 2 2
1 1
2 5
= 109677
21
100cm
–1
1 100
v 21 109677
= 4.34 10
–5 cm
The line belongs to Balmer series and lies in visible region of spectrum.
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CHAPTER 2: Structure of Atom
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2.57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often
used for the highly magnified images of biological molecules and other type of material. If the
velocity of the electron in this microscope is 1.6 106 ms
–1, calculate de Broglie wavelength
associated with this electron.
Ans: = ?, v = 1.6 106 m/s ; m = 9.1 10
–31 kg
34
6 31
h 6.6 10
mv 1.6 10 9.1 10
= 4.5 10
–10 m = 450 pm
2.58 Similar to electron diffraction, neutron diffraction microscope is also used for the determination of
the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic
velocity associated with the neutron.
Ans: Mass of neutron, m = 1.67 10–27
kg
= 800 pm = 800 10–12
m, v = ?
= h
mv or v =
34
27 12
h 6.6 10
m 1.67 10 800 10
= 494 m/s
2.59 If the velocity of the electron in Bohr’s first orbit is 2.19 106 ms
–1, calculate the de Broglie
wavelength associated with it.
Ans: 34
31 6
h 6.6 10
mv 9.1 10 2.19 10
= 3.31 10
–10 m = 331 pm
2.60 The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 105 ms
–1. If
the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with
this velocity.
Ans: 34
5
h 6.6 10
mv 0.1 4.37 10
= 1.51 10
–38 m
2.61 If the position of the electron is measured within an accuracy of 0.002 nm, calculate the uncertainty
in the momentum of the electron. Suppose the momentum of the electron is h/4 m 0.05 nm, is there
any problem in defining this value.
Ans: x = 0.002 nm = 0.002 10–9
m, p = ?
x . p = h
4
or 34
9
h 6.6 10p
4 x 4 3.14 0.002 10
= 2.62 10
–23
2.62 The quantum numbers of six electrons are given below. Arrange them in order of increasing
energies. If any of these combination(s) has/have the same energy lists:
1. n = 4, l = 2, ml = –2 , ms = –1/2
2. n = 3, l = 2, ml = 1 , ms = +1/2
3. n = 4, l = 1, ml = 0 , ms = +1/2
4. n = 3, l = 2, ml = –2 , ms = –1/2
5. n = 3, l = 1, ml = –1 , ms = +1/2
6. n = 4, l = 1, ml = 0 , ms = +1/2
Ans: (1) > (6) = (3) > (2) = (4) > (5)
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2.63 The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p
orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear
charge?
Ans: 4p orbital
2.64 Among the following pairs of orbitals which orbital will experience the larger effective nuclear
charge?
(i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.
Ans: (i) 2s (ii) 4d (iii) 3p
2.65 The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more
effective nuclear charge from the nucleus?
Ans: Si
2.66 Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Ans: (a) P (Z = 15) : 1s22s
22p
63s
23p
3, No. of unpaired electrons = 3
(b) Si (Z = 14) : 1s22s
22p
63s
23p
3, No. of unpaired electrons = 2
(c) Cr (Z = 24) : [Ar] 3d54s
1, No. of unpaired electrons = 6
(d) Fe (Z = 26) : [Ar] 3d64s
2, No. of unpaired electrons = 4
(e) Kr : No. of unpaired electrons in noble gases = 0
2.67 (a) How many sub-shells are associated with n = 4?
(b) How many electrons will be present in the sub-shells having ms value of –1/2 for n = 4?
Ans: (a) No. of sub–shells = n2 = 4
2 = 16
(b) 16 electrons