The photoelectric work function of potassium is 2.3 eV. If light having wavelength of 2800 `A^@` falls on potassium , find
the kinetic energy in electron volts of the most energetic electrons ejected.

Explanation

Given w = `2.3 e V, lambda = 2800 A^@`
`therefore` E ("in" e V)= (12375)/(lambda ("in" A^@)) = (12375)/(2800) = 4.4 e V`
`k_(max) = E-W = (4.4 - 2.3 ) eV = 2.1 eV`