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MHT CET · Physics · Dual Nature of Matter

Photoelectric emission takes place from a certain metal at threshold frequency ' \(v\) '. If the radiation of frequency ' \(2 v\) ' is incident on the metal plate, the maximum velocity of the emitted photoelectron will be ( \(\mathrm{m}=\) mass of electron, \(\mathrm{h}\) = Planck's constant)

  1. A \(\sqrt{\frac{2 \mathrm{hv}}{\mathrm{m}}}\)
  2. B \(\sqrt{\frac{\mathrm{hv}}{2 \mathrm{~m}}}\)
  3. C \(\sqrt{\frac{\mathrm{hv}}{3 \mathrm{~m}}}\)
  4. D \(\sqrt{\frac{\mathrm{hv}}{\mathrm{m}}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\sqrt{\frac{2 \mathrm{hv}}{\mathrm{m}}}\)

Step-by-step Solution

Detailed explanation

Einstein's photo electric equation
Concept: The maximum kinetic energy of photoelectron is written:
\(\frac{1}{2} \mathrm{mv}_{\mathrm{m}}^2 \leq \mathrm{hv}-\theta\)
The threshold frequency is given by
\(\mathrm{hv}_{\mathrm{c}}=\theta\)
Given, \(\mathrm{v}_{\mathrm{c}}=\mathrm{v}\), now for radiation with frequency \(2 \mathrm{v}\)
\(\begin{aligned}
& \frac{1}{2} \mathrm{mv}_{\mathrm{m}}^2 \leq \mathrm{h}(2 \mathrm{v}-\mathrm{h}(\mathrm{v})) \\
& \Rightarrow \mathrm{v}_{\mathrm{m}} \sqrt{\frac{2 \mathrm{hv}}{\mathrm{m}}}
\end{aligned}\)
Option (A) is correct.
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