Question

The photoelectric work - function of potassium is 2.3 eV. If light
having a wavelength of `2800 Å` falls on potassium, find
(a) the kinetic energy in electron volts of the most energetic electrons ejected.
(b) the stopping potential in volts.

Answer 1

Correct Answer - A::B
Given, `w = 2.3 eV, lambda = 2800 Å`
`:. E(in eV) = (12375)/(lambda(inÅ))=(12375)/(2800) = 4.4 eV` (a) `K_(max) = E - W = (4.4 - 2.3)eV
=2.1 eV` (b) `K_(max) = eV_0 2.1eV_0 or V_0 = 2.1 volt.