Question

The de Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency (v) of the incident radiation as, [$v_0$ is thrshold frequency] :

• A. $\lambda\propto \frac{1}{\left(v-v_{0}\right)^{\frac{3}{2}}}$
• B. $\lambda\propto \frac{1}{\left(v-v_{0}\right)^{\frac{1}{2}}}$
• C. $\lambda\propto \frac{1}{\left(v-v_{0}\right)^{\frac{1}{4}}}$
• D. $\lambda\propto \frac{1}{\left(v-v_{0}\right) } $

Answer 1

Answer: (B)

For electron
$\lambda_{DB} = \frac{\lambda}{\sqrt{2mK.E.}} $ (de broglie wavelength)
By photoelectric effect
$ hv = hv_{0} + KE $
$ KE = hv - hv_{0} $
$ \lambda_{DB} =\frac{h}{\sqrt{2m\times\left(hv -hv_{0}\right)}} $
$\lambda_{DB} \propto \frac{1}{\left(v-v_{0}\right)^{\frac{1}{2}} }$