Question

Radiation of frequency v=9v° is incident on a metal surface,having threshold frequency v° .The maximum velocity of the ejected photoelectrons is 6×10^6 m/s .what will be the maximum velocity of the photoelectrons if v is reduced to 3v°

Answer 1

given-

Radiation of frequency v=9v°
threshold frequency v°
maximum velocity of the ejected photoelectrons = 6×10^6 m/s

find-
maximum velocity of the photoelectrons if v° is reduced to 3v°

1/2 x m x (Vmax1)^2 = h(v1) - ϕ
1/2 x m x (6×10^6)^2 = 9hv° - hv°
1/2 x m x (6×10^6)^2 = 8hv° ------(i)

then
1/2 x m x (Vmax2)^2 = h(v2) - ϕ
1/2 x m x (Vmax2)^2 = 3hv° - hv°
1/2 x m x (Vmax2)^2 = 2hv° ------(ii)

eq (ii) / eq (i)
[(Vmax2)^2] / [(6×10^6)^2] = 1/4
 (Vmax2)^2 = (6×10^6)^2 / 4
(Vmax2) =  √[(6×10^6)^2 / 4]
Vmax2 = (6×10^6)^2 / 2 = 3 x 10^6 m/s


i hope it will help you
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