Question

a photosensitive surface has work function w°=hv°.If photon of energy 2hv° fall on this surface,the electrons are ejected with a maximum velocity of 4×10^6 m/s . When the energy of the incident photon is increased to 5hv°,then maximum velocity of the photoelectrons will be

a) 2×10^7 m/s. b)8×10^6 m/s

c)8×10^5 m/s. d)2×10^6 m/s

Answer 1

given-

Radiation of frequency v=2v°
threshold frequency v°
maximum velocity of the ejected photoelectrons = 4×10^6 m/s

find-
maximum velocity of the photoelectrons if v° is increased to 5v°

1/2 x m x (Vmax1)^2 = h(v1) - ϕ
1/2 x m x (4×10^6)^2 = 2hv° - hv°
1/2 x m x (4×10^6)^2 = hv° ------(i)

then
1/2 x m x (Vmax2)^2 = h(v2) - ϕ
1/2 x m x (Vmax2)^2 = 5hv° - hv°
1/2 x m x (Vmax2)^2 = 4hv° ------(ii)

eq (ii) / eq (i)
[(Vmax2)^2] / [(4×10^6)^2] = 4/1
 (Vmax2)^2 = (4×10^6)^2 x 4
(Vmax2) =  √[(4×10^6)^2 x 4]
Vmax2 = (4×10^6)^2 x 2 = 8 x 10^6 m/s


i hope it will help you
regards