Question

An electromagnetic wave of wavelength ‘λ’ is incident on a photosensitive surface of negligible work function. If ‘m’ mass is of photoelectron emitted from the surface that has de-Broglie wavelength λ_d, then

• A. \(\lambda=(\frac{2h}{mc})\lambda d^2\)
• B. \(\lambda=(\frac{2m}{hc})\lambda d^2\)
• C. \(\lambda_d=(\frac{2mc}{h})\lambda^2\)
• D. \(\lambda=(\frac{2mc}{h})\lambda d^2\)

Answer 1

The Correct Option is D

To solve this problem, we need to relate the given electromagnetic wave of wavelength \(\lambda\) with the de-Broglie wavelength \(\lambda_d\) of the emitted photoelectron. The key points to consider are:

1. The energy of the incident electromagnetic wave can be expressed using the formula for photon energy: \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant, and \(c\) is the speed of light. 
2. Since the work function is negligible, all the energy of the photon is converted into the kinetic energy of the emitted photoelectron.
3. The kinetic energy of the photoelectron can be related to its de-Broglie wavelength using the formula: \(\lambda_d = \frac{h}{mv}\), where \(m\) is the mass of the photoelectron and \(v\) is its velocity.

Now let's deduce the relation step-by-step:

1. From the energy conservation perspective, the kinetic energy (\(K.E.\)) of the emitted electron is given as: \(K.E. = \frac{1}{2}mv^2 = \frac{hc}{\lambda}\).
2. Using the de-Broglie relationship, we have: \(\lambda_d = \frac{h}{mv}\).
3. By rearranging the above de-Broglie equation for velocity \(v\), we get: \(v = \frac{h}{m\lambda_d}\).
4. Substitute the velocity expression back into the kinetic energy expression: \(\frac{1}{2}m \left(\frac{h}{m\lambda_d}\right)^2 = \frac{hc}{\lambda}\).
5. Simplifying the equation, we obtain: \(\frac{h^2}{2m\lambda_d^2} = \frac{hc}{\lambda}\).
6. Rearranging terms to solve for the incident wavelength \(\lambda\), we have: \(\lambda = \frac{2mc}{h} \lambda_d^2\).

Thus, the correct relation between the original wavelength \(\lambda\) and the de-Broglie wavelength \(\lambda_d\) is: \(\lambda = (\frac{2mc}{h})\lambda d^2\). This matches the given correct answer.