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 eV. If the stopping potential of the photoelectron is 10 V, then the value of n is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is C 4

$\frac{hc}{\lambda }=e{V}_0+{\varphi }_0=10eV+2.75eV=12.75eV$

But $\frac{hc}{\lambda }=13.6[\frac{1}{12}-\frac{1}{n^2}]eV\Rightarrow 1-\frac{1}{n^2}=\frac{12.75}{13.6}$

$\Rightarrow \frac{1}{n^2}=0.0625\Rightarrow n^2=\frac{10000}{625}=16\Rightarrow n=4$