Q144 / 200 · NEET 2006 · Physics

When photons of energy $h v$ fall on an aluminium plate (of work function $E_0$ ), photoelectrons of maximum kinetic energy $K$ are ejected. If the frequency of radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be:

  • Correct answer
    A
    $K+h v$
  • B
    $K+E_0$
  • C
    $2 K$
  • D
    $K$

Solution

According to given situation $\begin{aligned} & h v & =E_0+K \\ \text { and } & 2 h v & =E_0+k^{\prime} \\ \Rightarrow & k^{\prime} & =k+h v\end{aligned}$

Hence, the correct answer is option: 'A'.

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