Question

For photoelectric emission from certain metal the cut-off frequency is v. If radiation of frequency 2v impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)(a) $\sqrt{\frac{hv}{m}}$(b) $\sqrt{\frac{2hv}{m}}$(c) $2\sqrt{\frac{hv}{m}}$(d) $\sqrt{\frac{hv}{(2m)}}$

Answer 1

hey mate

the cprrect option is d

Answer 2

Answer:

√2hv / m

Explanation:

Cut off frequency = v (Given)

Radiation frequency = 2v (Given)

Work function, ϕ = hv  

According to the concept of einstein's photoelectric equation

1/2mv²max = h(2v)-hv

1/2mv²max = hv

hv’ = hv + Kmax

h.2v = hv + 1/2 mV²max [ v’ = 2v ]

hv = 1/2 mV²max = Vmax

= √2hv / m

Thus, the maximum possible velocity of the emitted electron will be √2hv / m