Q166 / 170 · Manipal (MET) 2016 · Physics

In a photoemissive cell with exciting wave length $\lambda$, the fastest electron has a speed $v$. If the exciting wavelength is changed to $3 \lambda / 4$, then the speed of the fastest emitted electron will be

  • A
    $\left(\frac{3}{4}\right)^{\frac{1}{2}}$
  • B
    $v\left(\frac{4}{3}\right)^{\frac{1}{2}}$
  • C
    less than $v\left(\frac{4}{3}\right)^{\frac{1}{2}}$
  • Correct answer
    D
    greater than $v\left(\frac{4}{3}\right)^{\frac{1}{2}}$

Solution

From Einstein's photoelectric equation, $\frac{1}{2} m v^2=\frac{h c}{\lambda}-W\ldots$ (i) and $\quad \frac{1}{2} m v^{\prime 2}=\frac{h c}{3 \lambda / 4}-W\ldots$ (ii) On dividing Eq. (ii) by Eq. (i), we get $\left(\frac{v^{\prime}}{v}\right)^2=\frac{\frac{4 h c}{3 \lambda}-W}{\frac{h c}{\lambda}-W}$ $\begin{array}{ll}\because & W \gt 0 \\ \therefore & v^{\prime} \gt v\left(\frac{4}{3}\right)^{\frac{1}{2}}\end{array}$

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

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