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: Write Einstein's photoelectric equation.
\[ K = h\nu - \phi \]
where \(K\) is the maximum kinetic energy, \(h\nu\) is the photon energy and \(\phi\) is the work function.
Step 2: Kinetic energy at the original frequency \(\nu\).
\[ K_1 = h\nu - \phi \]
Step 3: Kinetic energy when frequency is doubled to \(2\nu\).
\[ K_2 = h(2\nu) - \phi = 2h\nu - \phi \]
Step 4: Compare \(K_2\) with twice \(K_1\).
\[ 2K_1 = 2(h\nu - \phi) = 2h\nu - 2\phi \]
So \(K_2 - 2K_1 = (2h\nu - \phi) - (2h\nu - 2\phi) = \phi > 0.\)
Step 5: Since \(K_2\) exceeds \(2K_1\) by the positive work function \(\phi\), the new kinetic energy is somewhat more than double. It cannot be exactly double (that would need \(\phi = 0\)) nor four times.
\[\boxed{\text{somewhat more than double}}\]