Question

When a photon of frequency $v$ hits the surface of a metal with a threshold frequency $v_0(v>v_0)$ then which of the following options is correct about the ejected photoelectron?

• A. wavelength is $\frac{c}{v}$
• B. wavelength is $\frac{c}{v-v_0}$
• C. wavelength is proportional to $(v-v_0)$
• D. square of the wavelength is inversely proportional to $(v-v_0)$

Answer 1

The correct option is D square of the wavelength is inversely proportional to $(v-v_0)$

Energy of ejected electron:
$hv=h{v}_0+K.E.$
$\Rightarrow K.E.=h(v-v_0)$
where $v_0$ is threshold frequency.
Using de-Broglie equation, wavelength of photoelectron is given by
$\lambda =\frac{h}{\sqrt{2m\times K.E.}}$
$\lambda =\frac{h}{\sqrt{2m\times h(v-v_0)}}$

$\Rightarrow {\lambda }^2=\frac{h}{2m(v-v_0)}$