Question

In photoelectric emission, the frequency of incident light is doubled. Kinetic energy of emitted photoelectrons will become:

• A. double
• B. somewhat more than double
• C. less than double
• D. four times

Hint:

Use \(K = h\nu - \phi\); only the \(h\nu\) term doubles, \(\phi\) stays fixed.

Answer 1

The Correct Option is B

Step 1: Think in terms of what part of photon energy scales.
Only the photon energy \(h\nu\) scales with frequency; the work function \(\phi\) is a fixed property of the metal.
Step 2: Split the kinetic energy.
\(K = h\nu - \phi\). Doubling frequency doubles the \(h\nu\) term but leaves \(\phi\) unchanged, so the subtracted amount stays the same instead of also doubling.
Step 3: Because we subtract a smaller relative amount the second time, the result overshoots exactly double.
Numerically, if \(h\nu = 5\) eV and \(\phi = 2\) eV, then \(K_1 = 3\) eV and \(K_2 = 10 - 2 = 8\) eV, whereas double would be \(6\) eV.
Step 4: \(8 > 6\), confirming the energy becomes a little more than double.
\[\boxed{\text{somewhat more than double}}\]