Q82 / 120 · VITEEE 2016 · Physics

In photoelectric effect, initially when energy of electrons emitted is $E_{o}$, de-Broglie wavelength associated with them is $\lambda_{0}$. Now, energy is doubled then associated de-Broglie wavelength $\lambda^{\prime}$ is

  • Correct answer
    A
    $\lambda^{\prime}=\frac{\lambda_{o}}{\sqrt{2}}$
  • B
    $\quad \lambda^{\prime}=\sqrt{2} \lambda_{\mathrm{o}}$
  • C
    $\lambda^{\prime}=\lambda_{\mathrm{o}}$
  • D
    $\quad \lambda^{\prime}=\frac{\lambda_{\mathrm{o}}}{2}$

Solution

de-Broglie wavelength is given by $\begin{array}{l} \lambda=\frac{\mathrm{h}}{\mathrm{p}}, \text { where } \mathrm{h}=\text { Planck's constant and } \\ \mathrm{p}=\text { momentum } \end{array}$ Also, energy (E) and momentum are related as $\begin{array}{l} \mathrm{E}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}} \\ \therefore \quad \mathrm{p}=\sqrt{2 \mathrm{mE}} \end{array}$ $\therefore \quad \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}} \times \frac{1}{\sqrt{\mathrm{E}}}$ as $\mathrm{h}$ and $\mathrm{m}$ are constants Hence, $\frac{\lambda_{0}}{\lambda^{\prime}}=\sqrt{\frac{\mathrm{E}^{\prime}}{\mathrm{E}}}=\sqrt{\frac{2 \mathrm{E}}{\mathrm{E}}}=\sqrt{2}$ $\therefore \quad \lambda^{\prime}=\frac{\lambda_{0}}{\sqrt{2}}$

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

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