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

Options

A. 2
B. 3
C. 4
D. 5

Answer

C

Explanation

Here , stopping potential , `V_(0)= 10V`
Work function , `W=2.75e V`.
According to Einstein.s photoelectric equation
`e V_(0)=hv-W`
or `hv=Ev_(0)+W`
`=10eV +2.75eV =12.75eV`…..(i)
When an electron in the hydrogen atom makes a transition from excited state n to the ground state (n=1), then the frequency (v) of the emitted photon is given by
`hv=E_(n) -E_(1)`
`hv=-(13.6)/(n^2)-(-(13.6)/(1^2))`.
`[because "For hydrogen atom", E_(n) =-(13.6)/(n^2)eV]`
According to given problem
`(13.6)/(n^2)+13.6 =12.75`
`(13.6)/(n^2)=0.85 rArr n^(2) =(13.6)/(0.85) =16`
or n=4 .