Q155 / 160 · AP EAMCET 2023 (19 May Shift 1) · Physics

The de Broglie wavelength of a charged particle accelerated through a potential difference $\mathrm{V}$ is $\lambda$. If the potential difference is increased by $21 \%$, the de Broglie wavelength of the charged particle is

  • A
    $\frac{5 \lambda}{9}$
  • B
    $\frac{7 \lambda}{9}$
  • C
    $\frac{9 \lambda}{11}$
  • Correct answer
    D
    $\frac{10 \lambda}{11}$

Solution

De Broglie wavelength is given by $$ \begin{aligned} & \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}} \Rightarrow \lambda \propto \frac{1}{\sqrt{\mathrm{V}}} \\ & \mathrm{V}^{\prime}=\mathrm{V}+\frac{21}{100} \mathrm{~V}=1.21 \mathrm{~V} \\ & \frac{\lambda}{\lambda^{\prime}}=\frac{1}{\sqrt{\mathrm{V}}} \times \frac{\sqrt{\mathrm{V}^{\prime}}}{1} \\ & \frac{\lambda}{\lambda^{\prime}}=\sqrt{\frac{1.21 \mathrm{~V}}{\mathrm{~V}}}=\frac{11}{10} \Rightarrow \lambda^{\prime}=\frac{10}{11} \lambda \end{aligned} $$

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

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