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

To determine the correct expression for the de-Broglie wavelength \(\lambda_e\) of the emitted electrons from a photo-emissive substance, we need to consider the photoelectric effect and apply the concept of energy conservation. 

The energy of the incident photon can be given by:

\( E_{\text{incident}} = \frac{hc}{\lambda_i} \)

where \( \lambda_i \) is the wavelength of the incident radiation, \( h \) is the Planck's constant, and \( c \) is the speed of light.

According to the photoelectric effect, the energy of the incident photon is used to overcome the work function \(\phi\) of the substance and to provide kinetic energy to the emitted electron. Therefore, we have:

\( \frac{hc}{\lambda_i} = \phi + \frac{1}{2}mv^2 \)

where \( \frac{1}{2}mv^2 \) is the kinetic energy of the emitted electron with velocity \( v \) and mass \( m \).

The longest wavelength of radiation that can emit photoelectron is \( \lambda_0 \), which corresponds to the work function \(\phi\), so:

\( \phi = \frac{hc}{\lambda_0} \)

Substitute \(\phi\) into the equation for the incident energy:

\( \frac{hc}{\lambda_i} = \frac{hc}{\lambda_0} + \frac{1}{2}mv^2 \)

Simplify to find the expression for the kinetic energy:

\( \frac{1}{2}mv^2 = \frac{hc}{\lambda_i} - \frac{hc}{\lambda_0} \)

Now, the de-Broglie wavelength \(\lambda_e\) of the electron is given by:

\( \lambda_e = \frac{h}{mv} \)

To find \(v\), use the equation for kinetic energy:

\( \frac{1}{2}mv^2 = \frac{h \cdot c}{\lambda_i} - \frac{h \cdot c}{\lambda_0} \)

Solving for \(v\), we have:

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

Replacing \(v\) in the de-Broglie wavelength formula gives us:

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

Thus, the correct option is: \(<\lambda_e = \frac{h}{\sqrt{2mc \left(\frac{h}{\lambda_i} - \frac{h}{\lambda_0} \right)}}\).

Answer 2

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.