Question:
The de Broglie wavelength $(\lambda)$ associated with a photoelectron varies with the frequency $(v)$ of the incident radiation as, $\left[v_{0}\right.$ is threshold frequency $]$ :
Correct Option: , 4
Solution:
According to de-Broglie wavelength equation,
$\lambda=\frac{h}{m v} \Rightarrow \lambda \propto \frac{1}{v}$
According to photoelectric effect,
$\mathrm{h} v-\mathrm{h} v_{0}=\frac{1}{2} \mathrm{mv}^{2} ; v-v_{0}=\frac{1}{2} \frac{\mathrm{mv}^{2}}{\mathrm{~h}}$
$v-v_{0} \propto v^{2}$
$v \propto\left(v-v_{0}\right)^{1 / 2}$
$\therefore \lambda \propto \frac{1}{\left(v-v_{0}\right)^{1 / 2}}$