Q2 / 160 · AP EAMCET 2024 (22 May Shift 1) · Chemistry

If $v_0$ is the threshold frequency of a metal X , the correct relation between de Broglie wavelength $(\lambda)$ associated with photoelectron and frequency $(v)$ of the incident radiation is

  • Correct answer
    A
    $\lambda \propto \frac{1}{\sqrt{v-v_0}}$
  • B
    $\lambda \propto \frac{1}{\left(v-v_0\right)^{\frac{1}{4}}}$
  • C
    $\lambda \propto \frac{1}{\left(v-v_0\right)^{\frac{3}{4}}}$
  • D
    $\lambda \propto \sqrt{v-v_0}$

Solution

According to de Broglie wavelength, $\begin{aligned} & \lambda=\frac{\mathrm{h}}{\mathrm{mv}} \\ & \mathrm{KE}=\frac{1}{2} \mathrm{~m} v^2 \\ & v=\sqrt{\frac{2 \times \mathrm{K} \cdot \mathrm{E}}{\mathrm{~m}}} \\ & \therefore \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times \mathrm{K} \cdot \mathrm{E}}} \end{aligned}$
From photoelectric effect $\begin{aligned} & \mathrm{KE}=\mathrm{hv}-\mathrm{hv}_0 \\ & \mathrm{KE}=\mathrm{h}\left(\mathrm{v}-\mathrm{v}_0\right) \\ & \therefore \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mh}\left(v-v_0\right)}} \\ & \lambda=\sqrt{\frac{h}{2 \mathrm{~m}\left(v-v_0\right)}} \\ & \lambda \propto\left(\frac{1}{v-v_0}\right)^{1 / 2} \end{aligned}$

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

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