Q103 / 150 · MHT CET 2023 (10 May Shift 1) · Physics

An electron accelerated through a potential difference ' $V_1$ ' has a de-Broglie wavelength ' $\lambda$ '. When the potential is changed to ' $\mathrm{V}_2$ ' its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)$ is

  • A
    3:1
  • Correct answer
    B
    9:4
  • C
    3:2
  • D
    4:1

Solution

For electron, de Broglie wavelength, $\lambda=\frac{1.228}{\sqrt{\mathrm{V}}}$ Given: $\lambda_2=\lambda_1+0.5 \lambda_1=1.5 \lambda_1$ $\begin{aligned} & \therefore \quad \frac{\lambda_2}{\lambda_1}=\sqrt{\frac{V_1}{V_2}} \\ & \quad \Rightarrow \frac{V_1}{V_2}=\left(\frac{\lambda_2}{\lambda_1}\right)^2=\left(\frac{1.5 \lambda_1}{\lambda_1}\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4} \end{aligned}$

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

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