Atoms Ques 17
17. 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 $2.75 eV$. If the stopping potential of the photoelectron is $10 V$, the value of $n$ is
(a) 3
(b) 4
(c) 5
(d) 2
[2011M]
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Answer:
Correct Answer: 17.(b)
Solution:
- (b) $KE _{\max }=10 eV$
$\phi=2.75 eV$
Total incident energy
$E=\phi+KE _{\max }=12.75 eV$
$\therefore$ Energy is released when electron jumps from the excited state $n$ to the ground state.
$\because E_4-E_1={-0.85-(-13.6) ev}$
$=12.75 eV$
$\therefore$ value of $n=4$
BETA
