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 $]$ :

  1. $\lambda \propto \frac{1}{\left(v-v_{0}\right)}$

  2. $\lambda \propto \frac{1}{\left(v-v_{0}\right) \frac{1}{4}}$

  3. $\lambda \propto \frac{1}{\left(v-v_{0}\right)^{\frac{3}{2}}}$

  4. $\lambda \propto \frac{1}{\left(v-v_{0}\right)^{\frac{1}{2}}}$

Correct Option: , 4


According to de-Broglie wavelength equation,

$\lambda=\frac{h}{m v} \Rightarrow \lambda \propto \frac{1}{v}$

According to photoelectric effect,

$\mathrm{h} v-\mathrm{h} v_{0}=\frac{1}{2} \mathrm{mv}^{2} ; v-v_{0}=\frac{1}{2} \frac{\mathrm{mv}^{2}}{\mathrm{~h}}$

$v-v_{0} \propto v^{2}$

$v \propto\left(v-v_{0}\right)^{1 / 2}$

$\therefore \lambda \propto \frac{1}{\left(v-v_{0}\right)^{1 / 2}}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now