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

Step 1: Understanding the Photoelectric Effect.
The maximum kinetic energy \( K_{max} \) of the emitted electrons is given by Einstein's photoelectric equation: \[ K_{max} = \frac{hc}{\lambda_i} - \phi. \] The work function \( \phi \) is related to the threshold wavelength \( \lambda_0 \) by \( \phi = \frac{hc}{\lambda_0} \). Substituting this into the equation for \( K_{max} \): \[ K_{max} = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{\lambda_i} - \frac{1}{\lambda_0} \right). \quad \cdots (1) \] 
Step 2: Understanding de-Broglie Wavelength.
The de-Broglie wavelength \( \lambda_e \) of an electron with momentum \( p \) is given by \( \lambda_e = \frac{h}{p} \). 
The kinetic energy \( K_{max} \) of the emitted electron is related to its momentum \( p \) and mass \( m \) by \( K_{max} = \frac{p^2}{2m} \), so \( p = \sqrt{2m K_{max}} \). 
Substituting this into the de-Broglie wavelength equation: \[ \lambda_e = \frac{h}{\sqrt{2m K_{max}}}. \quad \cdots (2) \] 
Step 3: Combining the two equations.
Substitute the expression for \( K_{max} \) from equation (1) into equation (2): \[ \lambda_e = \frac{h}{\sqrt{2m \left( hc \left( \frac{1}{\lambda_i} - \frac{1}{\lambda_0} \right) \right)}}. \] 
Step 4: Simplifying the expression.
\[ \lambda_e = \frac{h}{\sqrt{2mch \left( \frac{1}{\lambda_i} - \frac{1}{\lambda_0} \right)}}. \] Looking closely at the options, option (1) is: \[ \lambda_e = \frac{h}{\sqrt{2mc \left( \frac{h}{\lambda_i} - \frac{h}{\lambda_0} \right)}} \] This can be simplified as: \[ \lambda_e = \frac{h}{\sqrt{2mch \left( \frac{1}{\lambda_i} - \frac{1}{\lambda_0} \right)}} \] This matches the derived expression. 
Therefore, option (1) is the correct answer.