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A Bohr's hydrogen atom undergoes a trans...

A Bohr's hydrogen atom undergoes a transition `n = 5 to n = 4` and emits a photon of frequency `f`. Frequency of circular motion of electron in `n = 4 orbit is f_(4)`. The ratio `f//f_(4)` is found to be `18//5 m`. State the value of `m`.

A

4

B

2

C

5

D

1

Text Solution

Verified by Experts

The correct Answer is:
C
`E_(n)=-(mZ^(2)e^(4))/(8epsilon_(0)^(2)n^(2)h^(2))`,
so, `hf=+(mZ^(2)e^(4))/(8epsilon_(0)^(2)h^(2))[1/16-1/25]`
`therefore f=(mZ^(2)e^(4))/(8epsilon_(0)^(2)h^(3))[9/(16 times 25)]" …(i)"`
and frequency `f_(4)=(Z^(2)e^(4)m)/(4epsilon_(0)^(2)n^(3)h^(3))=(Z^(2)e^(4)m)/(4epsilon_(0)^(2)(4)^(3)h^(3))" ...(ii)"`
`therefore f//f_(4)=18//25`, so m = 5
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