Question

If \( v_0 \) is the threshold frequency of a metal X, the correct relation between de Broglie wavelength \( \lambda \) associated with a photoelectron and frequency \( v \) of the incident radiation is:

• A. \( \lambda \propto \frac{1}{\sqrt{v - v_0}} \)
• B. \( \lambda \propto \frac{1}{(v - v_0)^{1/4}} \)
• C. \( \lambda \propto \frac{1}{(v - v_0)^{3/4}} \)
• D. \( \lambda \propto \sqrt{v - v_0} \)

Hint:

To determine the de Broglie wavelength of a photoelectron, use Einstein's photoelectric equation to find kinetic energy, then relate it to momentum.

Answer 1

The Correct Option is A

Step 1: Applying Einstein’s Photoelectric Equation The kinetic energy of the emitted photoelectron is: \[ K.E = h v - h v_0 \] Since kinetic energy is related to momentum \( p \) as: \[ K.E = \frac{p^2}{2m} \] Equating both expressions: \[ \frac{p^2}{2m} = h v - h v_0 \]
Step 2: Deriving de Broglie Wavelength From de Broglie’s equation: \[ \lambda = \frac{h}{p} \] Solving for \( p \): \[ p = \sqrt{2m (h v - h v_0)} \] Substituting into the de Broglie equation: \[ \lambda = \frac{h}{\sqrt{2m (h v - h v_0)}} \]
Step 3: Establishing Proportionality Since \( h \) and \( m \) are constants: \[ \lambda \propto \frac{1}{\sqrt{v - v_0}} \] Thus, the correct answer is: \[ \lambda \propto \frac{1}{\sqrt{v - v_0}} \]