When photons of energy hv fall on a metal plate of work function 'W0', photoelectrons of maximum kinetic energy 'K' are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

Question

When photons of energy hv fall on a metal plate of work function 'W0', photoelectrons of maximum kinetic energy 'K' are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be (A) K+W0 (B) K+hv (C) K (D) 2K

Tardigrade Admin · Accepted Answer

Given, energy of photon =h v, work function =W0 and maximum kinetic energy =K So, from the Einstein's photoelectric equation, E=K+W0 ⇒ h v=K+W0 ⇒ K=h v-W0...(i) If frequency of the radiation is doubled, then Einstein's photoelectric equation changed as K'=h(2 v)-W0...(ii) By substracting the Eq. (i) from (ii), we get K'-K=h(2 v)-h v-W0+W0 ⇒ K'=K+ h v So, option (b) is correct.

Q. When photons of energy $h v$ fall on a metal plate of work function $^{'} W_{0}^{'}$ , photoelectrons of maximum kinetic energy $^{'} K^{'}$ are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

$K + W_{0}$

$K + h v$

$K$

$2 K$

Q. When photons of energy hv fall on a metal plate of work function ′W0′​, photoelectrons of maximum kinetic energy ′K′ are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

K+W0​

K+hv

K

2K

Q. When photons of energy $h v$ fall on a metal plate of work function $^{'} W_{0}^{'}$ , photoelectrons of maximum kinetic energy $^{'} K^{'}$ are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be

$K + W_{0}$

$K + h v$

$K$

$2 K$