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प्रश्न
A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by
विकल्प
1.05 × 10−34 J s
2.11 × 10−34 J s
3.16 × 10−34 J s
4.22 × 10−34 J s
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उत्तर
1.05 × 10−34 J s
Let after absorption of energy, the hydrogen atom goes to the nth excited state.
Therefore, the energy absorbed can be written as
`10.2 = 13.6 xx (1/1^2 - 1/n^2)`
⇒ `10.2/13.6 = 1- 1/n^2`
`rArr = 1/(n^2) = (13.6 - 10.2)/13.6`
`rArr = 1/n^2 = 3.4/13.6`
`rArr n^2 = 4`
`rArr n = 2`
The orbital angular momentum of the electron in the nth state is given by
`L_n =(nh)/(2pi)`
Change in the angular momentum,
`DeltaL = (2h)/(2pi) - h/(2pi) = h/(2pi)`
`DeltaL = 1.05xx10^-34` Js
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संबंधित प्रश्न
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