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Question

An electron in the hydrogen atom jumps from excited state $n$ to the ground state. The wavelength so emitted illuminates a photosensitive material having work function $2.75 e V$ . If the stopping potential of the photoelectron is $10 V$ , the value of $n$ is

A

3

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B

4

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C

5

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D

2

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Solution

The correct option is C 4
${K E}_{m a x} = 10 e V$
$ϕ = 2.75 e V$
$E = ϕ + {K E}_{m a x} = 12.75 e V =$ Energy difference between $n = 4$ and $n = 1$
$\Rightarrow$ value of $n = 4$

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