Question

If the frequency of incident light falling on a photosensitive metal is doubled, the kinetic energy of the emitted photoelectron is

• A. unchanged
• B. halved
• C. doubled
• D. more than twice its initial value
• E. reduced to \(\frac{1}{4}\) th

Hint:

Doubling the frequency increases the total energy of the incoming photon. Since the "cost" to exit the metal (the work function) stays the same, all that extra energy goes directly into the electron's speed, making the final kinetic energy more than doubled.

Answer 1

The Correct Option is D

Concept: The Photoelectric Effect is governed by Einstein's equation.
• Einstein's Equation: \(K_{max} = h\nu - \Phi\), where \(K_{max}\) is the maximum kinetic energy, \(h\nu\) is the incident photon energy, and \(\Phi\) is the work function of the metal.
• Work Function: A constant value specific to the metal that represents the minimum energy required to eject an electron.

Step 1: Compare the initial and final states.
Initial kinetic energy: \(K_1 = h\nu - \Phi\). Final kinetic energy when frequency is doubled (\(2\nu\)): \[ K_2 = h(2\nu) - \Phi = 2h\nu - \Phi \]

Step 2: Relate \(K_2\) to \(K_1\).
We can rewrite \(K_2\) to see its relationship with \(K_1\): \[ K_2 = 2(h\nu - \Phi) + \Phi \] \[ K_2 = 2K_1 + \Phi \] Since the work function \(\Phi\) is a positive value, the new kinetic energy (\(K_2\)) is equal to twice the initial energy plus an additional constant amount. Therefore, \(K_2 > 2K_1\).