Question

In case of photoelectric emission from certain metal, the cutoff frequency is \(ν\). If the radiation of frequency \(3ν\) is incident on the metal plate, the maximum possible velocity of the emitted electrons will be (\(m\) = mass of electron, \(h\) = Planck's constant)

• A. \(\sqrt{\frac{hν}{2m}}\)
• B. \(\sqrt{\frac{hν}{m}}\)
• C. \(2(\sqrt{\frac{hν}{m}})\)
• D. \(\sqrt{\frac{2hν}{m}}\)

Hint:

Use Einstein equation with the work function equal to h times the cutoff frequency.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Einstein photoelectric equation: \(h f = \phi + K_{max}\), with \(\phi = h\nu_0\). The cutoff frequency here is \(\nu\).

Step 2: Key Formula or Approach:
\(K_{max} = h(3\nu) - h\nu = 2h\nu\), and \(K_{max} = \frac12 mv^2\).

Step 3: Detailed Explanation:
\(\frac12 mv^2 = 2h\nu\), so \(v^2 = \frac{4h\nu}{m}\).
\[ v = 2\sqrt{\frac{h\nu}{m}} \]
Option D, \(\sqrt{\frac{2h\nu}{m}}\), would correspond to a kinetic energy of \(h\nu\) only, which holds if the incident frequency were \(2\nu\).

Final Answer:
The maximum speed is \(2\sqrt{\frac{h\nu}{m}}\), option (C). \[ \boxed{2\sqrt{\frac{h\nu}{m}}} \]