Question

The light of wavelength \( \lambda \) incident on the surface of metal having work function \( \phi \) emits the electrons. The maximum velocity of electrons emitted is

• A. \( \frac{2(hc - \phi)}{m \lambda} \)
• B. \( \left( \frac{2(hc - \lambda \phi)}{m \lambda} \right)^{1/2} \)
• C. \( \frac{2(hc - \lambda)}{m \lambda} \)
• D. \( \frac{2(hv - \phi) \lambda}{mc} \)

Hint:

In the photoelectric effect, the maximum velocity of the emitted electrons depends on the wavelength of light and the work function of the material.

Answer 1

The Correct Option is B

Step 1: Understanding the photoelectric effect.
The energy of the incoming photon is \( E = hc / \lambda \), where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the incident light. The work function of the metal is \( \phi \), and the energy required to emit an electron is equal to the energy of the photon minus the work function.
Step 2: Maximum velocity of the electron.
The maximum kinetic energy of the emitted electron is given by the equation: \[ K.E. = \frac{1}{2} m v^2 = \frac{hc}{\lambda} - \phi \] Solving for the velocity \( v \), we get: \[ v = \left( \frac{2(hc - \lambda \phi)}{m \lambda} \right)^{1/2} \]
Step 3: Conclusion.
The maximum velocity of the emitted electron is \( \left( \frac{2(hc - \lambda \phi)}{m \lambda} \right)^{1/2} \), so the correct answer is (B).