Question

In photoelectric effect an em-wave is incident on a metal surface and electrons are ejected from the surface. If the work function of the metal is 2.14 eV and stopping potential is 2V, what is the wavelength of the em-wave?

(Given hc = 1242 eVnm where h is the Planck's constant and c is the speed of light in vaccum.)

(1) 400 nm

(2) 600 nm

(3) 200 nm

(4) 300 nm

Answer 1

Correct option is : (4) 300 nm

\(\mathrm{eV}_{\mathrm{s}}=\mathrm{E}-\Phi\) 

\( 2 \ \mathrm{eV}=\mathrm{E}-2.14 \ \mathrm{eV}\) 

\( \mathrm{E}=4.14 \ \mathrm{eV}\)

\( \mathrm{E}=\frac{\mathrm{hc}}{\lambda}\) 

\( \lambda=\frac{1242}{4.14}=300 \mathrm{~nm}\)