18-atoms-molecules-and-nuclei.pdf

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18. ATOMS, MOLECULES AND NUCLEI1. Find the ratio of diameter of electron in

1st Bohrs orbit to that in 4th bohrs orbit.Given :

r n = radius of nth Bohrs orbit,

d n = diameter of nth Bohrs orbit

To Find : d 1 : d4 = ?Formula :

rn =h nme

2 20

2

Solution :Sinced n = 2 rnFrom formula

d n =2 h n

me

2 20

2

d 1 =( )2 h 1

me

220

2

=2 h

me

20

2

d 4 = ( )2 h 4

me

220

2

= 162 h

me

20

2

dd

1

4=

2 hme

2 h16me

20

2

20

2

=1

16

d 1 : d 4 = 1 : 16

2. Calc ul at e the ch an ge in a ng ularmomentum of electron when it jumpsfrom 3 rd orbit to 1 st orbit in hydrogen

Given :n1 = 1 (first orbit)n2 = 3 (third orbit)

To Find : L3 L1 = ?Formula :

L = mvr =nh2

Solution :From formula

L1 = mv 1r1 =n h21

L3 = mv 3r3 =n h23

L3 L1 = mv 3r3 mv 1r1

= n h23 n h2

1

=h

2 (n 3 n 1)

=h

2 (3 1) =h

=6.63 10

3.142

34

L3 L1 = 2.11 1034 Js or kg m 2/s

3. Find the longest wavelength in Paschenseries. [Given R = 1.097 107 m 1]

Given :p = n 1 = 3n = n 2 = 4

To Find :

L = ?

Atoms, Molecules and Nuclei


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4. Find the f requency of revolution of

electron in 2nd

Bohrs orbit, if the radiusand the speed of electron in that orbitare 2.14 1010 m and 1.09 106 m/srespectively.

Given :r2 = 2.14 10

10 mn = 2v2 = 1.09 10

6 m/sTo Find :

v2 = ?Formula :

= v2 rSolution :

From formula

2 =

v2 r

2

2

=1.09 10

2 3.142 2.14 10

6

10

2 = 8.11 10

14 Hz

5. Find the value of Rydbergs constant, if

the energy of electron in second orbitin hydrogen atom is 3.4 eV.Given :

E2 = 3.4 eV= 3.4 1.6 1019 J

n = 2To Find :

R = ?Formula :

En = me

8 h n

4

2 2 20

Solution :

E2 = ( )me

8 h 2

4

22 20

[ n = 2]

E2 = ( )me

8 h 4

4

2 20

hchc

E2 =me

8 h c

4

2 30

hc4

But

me

8 h c

4

2 30

= R

E2 = Rhc4

R = 4Ehc

2

=( )4 3.4 1.6 10

6.63 10 3 10

19

34 8

=4 3.4 1.6 10

6.63 3 10

19

26

R = 1.094 107 m 1

Formula :

1L

= R 1 1n n

2 21 2


1

L= R

1 13 4

2 2

1L

= R1 19 16

= R1699 16

=1.097 10 7

9 16

7

L =

9 161.097 7

107

= 18.752 107 m

L = 18750 A.U

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1

L = R1699 16

1

L=

7R9 16

L =

9 167R

L

S

=

9 167R

R9

L

S

=

16

9

LS

= 2.286 : 1

8. The moving electron and a photon hassame deBroglie wavelength. Show thatthe electron possesses more energy thancarried by the photon.

Solution :Let,

= wavelength of moving electron= de Broglie wavelength of photon

Ee = Energy of high speed electronEp = Energy of photon of same wavelength

To Show that :Ee > Ep

Proof :de Broglie wavelength,

=h

mc

mc =h

p =h

Ep = h

=hc = pc ...(i)

where p = momentum which is same forelectron and photon.Rest mass of electron is given by

m =m

v1 c

02

2

where v = speed of electron c = speed of light

m2 =m

v1 c

20

2

2

m 2 (c2 v2) = m c2 20

m 2c2 m2v2 = m c2 20 ...(ii)Multiply equation (ii) by c 2

m 2c4 m2v2c2 = m c2 40(mc 2)2 (mv) 2c2 = (m 0c

2)2

(mc 2)2 = (m 0c2)2 (mv) 2c2 ...(iii)

But mc 2 = Ee,m0c

2 = E0,mv = pEquation (iii) becomes

E2e = E20 + p 2c2

Ee = E + p c2 2 20 ...(iv)

From equation (i) and (iv)Ee > Ep

9. A monochromatic light of wavelength is incident on hydrogen atom that liftsit to 3 rd orbit from ground level. Findthe wavelngth and frequency ofincident photon.

(Given : E 3 = 1.51 eV, E 1 = 13.6 eV)Given :E3 = 1.51 eVE1 = 13.6 eV

To Find : = ?

= ?Formula :

i) E =hcE


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.. 106 MAHESH TUTORIALS SCIENCE


ii) E = h

Solution :From formula (i)

=hcE

=hc

E E3 1

=( )hc

1.51 13.6 eV

=hc

12.08 1.6 1019

=6.63 10 3 10

12.08 1.6 10

34 8

19

= 1.029 107

= 1029 Ao

From formula (ii)

=E

h

= 12.08 1.6 106.63 10

1934

= 2.91 1015 Hz

10. An electron in hydrogen atom stays in2nd orbit for 10 8 s. How many,revolutions will it make till it jumps tothe ground state ?

Given :t = 108 s

To Find :

number of revolution v

= ?Formula : E = h


=E

h

=

E Eh

2 1

=( )3.4 13.6

6.63 1034

= 10.2 1.6 106.63 10

19

34

= 2.46 1015 HzIt means that electrons takes 1 s tocomplete 2.46 1015 revolutionIn 10 8 s electron will complete2.46 1015 108 revolutionv = 2.46 107 revolution

11. Find momentum of the electron having de Broglie wavelength of 0.5 A

o.

Given :de Broglie wavelength, = 0.5 A

o

= 0.5 1010 mTo Find :

p = ?Formula :

=hp

Solution :

From formula

p =h

=6.63 100.5 10

34

10

p = 13.26 1024 kg ms 1

12. Find the wavelength of a protonaccelerated by a potential difference of50 V. [Given m p = 1.673 10

27 kg]Given :

V = 50 voltmp = 1.673 10

27 kgTo Find :

= ?Formula :

=h

2m eVp


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Solution :

From formula

=6.63 10

2 1.673 10 1.6 10 50

34

27 19

=44

6.63 10

1.673 1.6 10

34

=6.63 10 10

1.673 1.6

34 22

= 0.04052 1010 m = 0.04 A.U

13. A cracker of mass M at rest explodes intwo parts of masses m 1 and m 2 with non-zero velocities. Find the ratio of the deBroglie wavelength of two particles.

Solution :Let

1 = de Broglie wavelength of mass m 1

2 = de Broglie wavelength of mass m 2

Initial momentum of mass m = p = 0

p1 = momentum of mass m 1p2 = momentum of mass m 2

From principle of conservation of momentum

p

= p + p1 2

0 = p + p1 2

p1 = p2

p1

= p2

... (i)

de Broglie wavelength is given by =hp

1 =

hp 1

and

2 =

hp 2

p1 =h

1 and ... (ii)

p2 =h

2... (iii)

From equations (i), (ii) and (iii)

h

1=

h

2

1

2= 1

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18-atoms-molecules-and-nuclei.pdf