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.

Options

A. `1.6 xx 10^(-19) J`
B. `16 xx 10^(10) J`
C. `3.2 xx 10^(-19) J`
D. `6.4 xx 10^(-10) J`

Answer

C

Explanation

KE = ("Energy of radiation= Work function")
`= (h xx (c )/(lambda) - 4.2)`
`= ((6.6 xx 10^(-34) J s xx 3 xx 10^(8) M)/(2000 xx 10^(-10)m)) - (4.2 xx 1.602 xx 10^(-19) J`
`= (9.9 xx 10^(-19) J) - (6.7 xx 10^(-19)J = 3.2 xx 10^(-19) J`