JEE MAIN - Physics (2025 - 7th April Evening Shift - No. 2)

A photoemissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_e$. The longest wavelength of radiation that can emit photoelectron is $\lambda_o$. Expression for de-Broglie wavelength is given by:

(m: mass of the electron, h: Planck's constant and c: speed of light)

$\lambda_e = \frac{\sqrt{h \lambda_i}}{\sqrt{2mc}}$
$\lambda_e = \frac{h}{\sqrt{2mc \left( \frac{1}{\lambda_i} - \frac{1}{\lambda_o} \right)}}$
$\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_{\mathrm{o}}}\right)}}$
$\lambda_e=\sqrt{\frac{h \lambda_0}{2 m c}}$

Explanation

Kinetic Energy (K.E.): The kinetic energy of the emitted electron is given by subtracting the work function (W) of the material from the energy of the incident radiation (E).

$ \text{K.E.} = E - W $

Expressions for Energy:

The energy of the incoming radiation $ E $ is calculated using the formula:

$ E = \frac{hc}{\lambda_i} $

The work function $ W $, which is the minimum energy needed to release an electron, when the radiation wavelength is at its longest $\lambda_0$, is:

$ W = \frac{hc}{\lambda_0} $

De-Broglie Wavelength of the Electron:

The de-Broglie wavelength $\lambda_e$ of the emitted electron is given by:

$ \lambda_e = \frac{h}{\sqrt{2m \cdot \text{K.E.}}} $

Substituting the Values:

The kinetic energy of the emitted electron can be expressed in terms of energies as:

$ \frac{h^2}{2m \lambda_e^2} = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} $

Final Expression for $\lambda_e$:

Solving for $\lambda_e$, the de-Broglie wavelength of the electron, we have:

$ \lambda_e = \sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}} $

This expression determines the de-Broglie wavelength of the emitted electron based on the wavelengths of the incident and threshold radiation.

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