Question

If the frequency of incident light falling on a metallic surface is doubled, maximum kinetic energy of emitted photoelectrons

• A. increases to slightly more than double
• B. remains the same
• C. becomes twice
• D. decreases to slightly more than double

Hint:

Work function reduces the doubling effect when frequency is increased.

Answer 1

The Correct Option is A

Step 1: Photoelectric equation.
\[ K_{\max} = h\nu - \phi \] where $\phi$ is the work function.

Step 2: Double the frequency.
New kinetic energy:
\[ K' = h(2\nu) - \phi = 2h\nu - \phi \]

Step 3: Compare with original kinetic energy.
\[ 2K_{\max} = 2(h\nu - \phi) = 2h\nu - 2\phi \] Since $\phi$ is constant,
\[ K'>2K_{\max} \]

Step 4: Conclusion.
The maximum kinetic energy increases to slightly more than double.