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