Question

When photons of energy $u$ fall on an aluminium plate, 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. $K+E_0$
• B. $2K$
• C. $K$
• D. $K+hv$

Answer 1

The correct option is D $K+hv$

Using Einstein equation of photoelectric effect
$W={\varphi }_0+KE$
$hv={\varphi }_0+K$
${\varphi }_0=hv-K$ .....(1)
If frequency of incident is double
$2hv={\varphi }_0+K_1$ ....(2)
By using equation (1)
$2hv=(hv-K)+K_1$ $({\varphi }_0=hv-K)$
$K1=hv+K$