Question

A photo-emissive 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_0 $. Expression for de-Broglie wavelength is given by : (m : mass of the electron, h : Planck's constant and c : speed of light)

• A. \( \lambda_e = \frac{h}{\sqrt{2mc \left( \frac{h}{\lambda_i} - \frac{h}{\lambda_0} \right)}} \)
• B. \( \lambda_e = \sqrt{\frac{h \lambda_0}{2mc}} \)
• C. \( \lambda_e = \frac{h}{\sqrt{2mch \left( \frac{1}{\lambda_i} - \frac{1}{\lambda_0} \right)}} \)
• D. \( \lambda_e = \sqrt{\frac{h \lambda_i}{2mc}} \)

Hint:

Remember the fundamental equations for the photoelectric effect (\(K_{max} = hf - \phi\)) and de-Broglie wavelength (\(\lambda = \frac{h}{p}\)). Relate the kinetic energy of the electron to its momentum to connect these two concepts. Pay close attention to algebraic manipulations to match the given options.

Answer 1

The Correct Option is A

The de-Broglie wavelength \(\lambda_e\) of emitted electrons from a photo-emissive substance is determined using the photoelectric effect and energy conservation. The incident photon energy is \( E_{\text{incident}} = \frac{hc}{\lambda_i} \), where \( \lambda_i \) is the incident radiation wavelength, \( h \) is Planck's constant, and \( c \) is the speed of light.

In the photoelectric effect, incident photon energy equals the work function \(\phi\) plus the kinetic energy of the emitted electron: \(\frac{hc}{\lambda_i} = \phi + \frac{1}{2}mv^2\), where \( \frac{1}{2}mv^2 \) is the kinetic energy of an electron with velocity \( v \) and mass \( m \).

The threshold wavelength \(\lambda_0\) corresponds to the work function: \(\phi = \frac{hc}{\lambda_0}\).

Substituting \(\phi\) into the incident energy equation yields: \(\frac{hc}{\lambda_i} = \frac{hc}{\lambda_0} + \frac{1}{2}mv^2\).

Rearranging for kinetic energy: \(\frac{1}{2}mv^2 = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0}\).

The de-Broglie wavelength \(\lambda_e\) of the electron is given by: \(\lambda_e = \frac{h}{mv}\).

From the kinetic energy equation: \(\frac{1}{2}mv^2 = \frac{h \cdot c}{\lambda_i} - \frac{h \cdot c}{\lambda_0}\).

Solving for \( v \): \( v = \sqrt{\frac{2hc}{m} \left(\frac{1}{\lambda_i} - \frac{1}{\lambda_0} \right)} \).

Substituting \( v \) into the de-Broglie wavelength formula: \(\lambda_e = \frac{h}{\sqrt{2mc \left(\frac{h}{\lambda_i} - \frac{h}{\lambda_0} \right)}} \).

The correct expression is: \(<\lambda_e = \frac{h}{\sqrt{2mc \left(\frac{h}{\lambda_i} - \frac{h}{\lambda_0} \right)}}\).