Answer:
(c) Both A and B
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
[tex]E = \frac{hc}{\lambda} \\\\E = \frac{(6.626*10^{-34})(3*10^8)}{3500*10^{-10}}\\\\E = 5.68 *10^{-19} \ J\\\\E = 5.68 \ eV[/tex]
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