Q158 / 180 · COMEDK 2012 · Physics

The graph of kinetic energy of photoelectron versus frequency of incident radiation is shown for two metals $M$ and $N$. We may definitely conclude
Figure

  • Correct answer
    A
    work function of $M>$ work function of $N$
  • B
    work function of $M < $ work function of $N$
  • C
    work function of $M=$ work function of $N$
  • D
    at the threshold frequency of $M$ the kinetic energy of the photoelectron emitted by $M$ is more than that emitted by $N$

Solution

Work function of a metal, $$ \begin{aligned} W &=h v_{0} \\ \text { where, } h &=\text { Planck's constant } \\ v_{0} &=\text { threshold frequency. } \end{aligned} $$ As, the threshold frequency of metal $M$ is more than that of metal $N$, so work function of $M$ is more than that of $N$, while their kinetic energies are equal.

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

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