Question

When photons of energy h fall on an aluminium plate (of work function E${}_0$), photoelectrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, then the maximum kinetic energy of the ejected photoelectrons will be:

• A. K + E${}_0$
• B. 2K
• C. K
• D. K + h

Answer 1

The correct option is D K + h

According to Einstein's photoelectric equation,

Energy of proton = $KE$ or protoelectron +work function of metal.

$hV=\frac{1}{2}m{V}^2+E_0$ --------(1)

Now, $V^{{}^{\prime }}=2V$

$K_{max}^{{}^{\prime }}=h\left(2V\right)-E_0$

$K_{max}^{{}^{\prime }}=2hV-E_0$ --------(2)

Then equation $1$ and equation $2$

$K_{max}^{{}^{\prime }}=2(K_{max}+E_0)-E_0$

$=2k\text{max}+E_0$

$=K_{max}+(K_{max}+E_0)$

$K_{max}+hV$ from equation $2$

Pulling $K_{max}=K$

$K_{max}^{{}^{\prime }}=K+hV$.