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 understand the relationship between the wavelengths of the incident electromagnetic wave and the de-Broglie wavelength of the emitted photoelectron. The problem involves Photoelectric Effect and de-Broglie Hypothesis.

Conceptual Understanding:

• The Photoelectric Effect explains that when light of sufficient frequency (or energy) strikes a material, it emits electrons. The energy of the incident photon should be sufficient to overcome the work function of the material.
• The de-Broglie Hypothesis states that every moving particle or object has an associated wave. For a particle like an electron, the de-Broglie wavelength \( \lambda_d \) is given by: \(\lambda_d = \frac{h}{p}\), where \( p \) is the momentum of the particle.

Solution Steps and Equations:

1. The energy of the incident photon with wavelength \( \lambda \) is given by: \(E = \frac{hc}{\lambda}\), where \( h \) is Planck's constant and \( c \) is the speed of light.
2. As the work function is negligible, all the energy of the photon goes into the kinetic energy of the emitted electron. Thus, the kinetic energy \( K \) is: \(K = \frac{hc}{\lambda}\).
3. The kinetic energy is also related to the de-Broglie wavelength by the relation: \(K = \frac{p^2}{2m}\), where \( p \) is the momentum and \( m \) is the mass of the electron.
4. From the de-Broglie relation \(\lambda_d = \frac{h}{p}\), we can express the momentum as: \(p = \frac{h}{\lambda_d}\).
5. Substitute for \( p \) in the kinetic energy relation: \(K = \frac{\left(\frac{h}{\lambda_d}\right)^2}{2m} = \frac{h^2}{2m\lambda_d^2}\).
6. Equating the two expressions for kinetic energy: \(\frac{hc}{\lambda} = \frac{h^2}{2m\lambda_d^2}\).
7. Solving for \( \lambda \): \(\lambda = \frac{2mc}{h} \lambda_d^2\).

Thus, the correct relation consistent with the de-Broglie hypothesis and the photoelectric effect is: \(\lambda=(\frac{2mc}{h})\lambda d^2\).

Conclusion: This equation correctly relates the wavelength of the incident electromagnetic wave, the de-Broglie wavelength, and other constants. Thus, the correct answer is option: \(\lambda=(\frac{2mc}{h})\lambda d^2\).