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Question:

An electron in the hydrogen atom jumps from an excited state 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, what is the value of n?

2

3

5

4

Solution:

KEmax=10eV
φ=2.75eV
E=φ+KEmax=12.75eV=Energy difference between n=4 and n=1 ⇒ value of n=4