Question

Light of wavelength 3500Å is incident on two metals A and B whose work functions are 3.2 eV and 1.9 eV respectively. Which metal will emit photoelectrons?

(a) A
(b) B
(c) Both A and B
(d) Neither A nor B

Answer 1

(c) Both A and B

Step-by-step explanation:

Given;

wavelength of the incident light, λ = 3500 Å = 3500 x 10⁻¹⁰ m

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f is frequency = c / λ

`E = (hc)/(\lambda) \\\\E = ((6.626*10^(-34))(3*10^8))/(3500*10^(-10))\\\\E = 5.68 *10^(-19) \ J\\\\E = 5.68 \ eV`

Work function is the minimum amount of energy required to liberate electrons from a metal surface.

The work function of metal A is 3.2 eV and

Th e work function of metal B is 1.9 eV

Both work functions are less than the incident energy of the light calculated as 5.68eV.

Thus, both metals will emit photoelectrons.

(c) Both A and B