Question

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

• A. $\lambda \propto \frac{1}{(v-v_0{)}^{\frac{3}{2}}}$
• B. $\lambda \propto \frac{1}{(v-v_0{)}^{\frac{1}{2}}}$
• C. a$\lambda \propto \frac{1}{(v-v_0{)}^{\frac{1}{4}}}$
• D. $\lambda \propto \frac{1}{(v-v_0)}$

Answer 1

The correct option is B $\lambda \propto \frac{1}{(v-v_0{)}^{\frac{1}{2}}}$

For electron

${\lambda }_{DB}=\frac{\lambda }{\sqrt{2mK.E.}}$ (de broglie wavelength)

By photoelectric effect

$hv=h{v}_0+KE$
${\lambda }_{DB}=\frac{h}{\sqrt{2m\times (hv-h{v}_0)}}$
${\lambda }_{DB}\alpha \frac{1}{(v-v_0{)}^{\frac{1}{2}}}$

So, the correct option is $B$