The electron in the hydrogen atom jumps from excited state (n=3) to its ground state (n=1) and the photons thus emitted irradiate a photosensitive material. If the work function of the material is 5.1eV, the stopping potential is estimated to be: (The energy of the electron in nth state is `E_(n)=-13.6//n^(2)eV`)

Options

A. 5.1V
B. 12.1V
C. 17.2 V
D. 7V

Answer

D

Explanation

`E_(3)- E_(1)= - 13.6 [(1)/((3)^(2))- (1)/(1^(2))] eV`
`= (13.6 x 8)/(9)= 12.1 eV`
Since, `E= eV rArr V= 12.1V`
Stoping potential `(V_(0)) = V- (phi_(0))/(e )` [`phi_(e) rarr` work function ]
`= 12.1 - 5.1 = 7V`