CBSE Class 12 Physics 2019 Outside Delhi Set 2 Paper

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Question : 6
Total: 9
SECTION -C

(a) How is the stability of hydrogen atom in Bohr model explained by de-Broglie's hypothesis?
(b) A hydrogen atom initially in the ground state absorbs a photon which excites it to n=4 level. When it gets de-excited, find the maximum number of lines which are emitted by the atom. Identify the series to which these lines belong. Which of them has the shortest wavelength?
Solution:  
(a) Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates. The second postulates is.

Electron revolves around the nucleus only in those orbits for which angular momentum in integral multiple of
h
2
.
de-Broglie had proposed that material particle such as electrons also have a wave nature. He argued that the electron in its circular orbit, as proposed by Bohr, must be seen as a particle wave. Drawing an analogy with waves travelling on the string, particle waves too can lead to formation of standing waves. In a string, standing waves are formed, when the total distance travelled by a wave back and forth is one wavelength, two wavelength or integral multiple of wavelengths. Other waves interfere with themselves after reflection and their amplitude falls to zero. For an electron moving in nth orbit with radius rn, its circumference is 2πrn

2πrn=nλ,n=1,2,3
From de-Broglie's hypothesis,
Wavelength of the election (λ) is given as
λ=
h
p
=
h
mv

For nth orbit, λ=
h
mvn

2πrn=
nh
mvn
or
mvnrn
=
nh
2

This is the second postulate of Bohr. That give the discrete orbits and energy levels in hydrogen atom. Thus de-Broglie explained the postulate of quantisation angular momentum.
(b) For ground state n=1 For de-excitation from n= 4 to n=1 we get spectral lines constituting Lyman series whose wavelength is given by the formula,
1
λ
=R(
1
12
1
n2
)

Here R= Rydberg constant number n=2,3,4
1
λ2
=R(1
1
4
)
=
3R
4
λ2
=
4
3R

1
λ3
=R(
1
12
1
32
)
=R(1
1
9
)

λ3=
9
8R

1
λ4
=R(
1
12
1
42
)
=R(1
1
16
)

λ4=
16
15R
[ Here,
1
R
=912
]
There would be maximum three lines emitted by the atom.
λ4 has the shortest wavelength.
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