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)

<p><p><strong>Kinetic Energy (K.E.):</strong> 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).</p> <p>$ \text{K.E.} = E - W $</p></p> <p><p><strong>Expressions for Energy:</strong></p></p> <p><p>The energy of the incoming radiation $ E $ is calculated using the formula:</p> <p>$ E = \frac{hc}{\lambda_i} $</p></p> <p><p>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:</p> <p>$ W = \frac{hc}{\lambda_0} $</p></p> <p><strong>De-Broglie Wavelength of the Electron:</strong></p> <p><p>The de-Broglie wavelength $\lambda_e$ of the emitted electron is given by:</p> <p>$ \lambda_e = \frac{h}{\sqrt{2m \cdot \text{K.E.}}} $</p></p> <p><strong>Substituting the Values:</strong></p> <p><p>The kinetic energy of the emitted electron can be expressed in terms of energies as:</p> <p>$ \frac{h^2}{2m \lambda_e^2} = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} $</p></p> <p><strong>Final Expression for $\lambda_e$:</strong></p> <p><p>Solving for $\lambda_e$, the de-Broglie wavelength of the electron, we have:</p> <p>$ \lambda_e = \sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0}\right)}} $</p></p> <p>This expression determines the de-Broglie wavelength of the emitted electron based on the wavelengths of the incident and threshold radiation.</p>