Question

Light of wavelength 4000 Å is incident
on a metal surface of work function 2.0
EV. The stopping potential will be
Value of n is
v.
10​

Answer 1

Note :- See answer from website.

Given that;-

Wavelength = $\large\rm { \lambda = 4000 \r{A} = 400 nm}$

$\large\rm { \phi = 2 \ eV}$

Solution:-

Wavelength = $\large\rm { \lambda = 4000 \r{A} = 400 nm}$

Energy of incident photon = $\large\rm { E = \frac{hc}{ \lambda} }$

$\large\rm { E = 1240/400 = 3.1 eV}$

Now,

$\large\rm { E - \phi}$

$\large\rm { 3.1-2 = \boxed{1.1 \rm{eV}}}$ is your answer.