Q102 / 200 · NEET 2024 (Re-NEET) · Physics

An electron and an alpha particle are accelerated by the same potential difference. Let $\lambda_{\varepsilon}$ and $\lambda_a$ denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:

  • Correct answer
    A
    $\lambda_e\gt\lambda_\alpha$
  • B
    $\lambda_e=4 \lambda_\alpha$
  • C
    $\lambda_e=\lambda \alpha$
  • D
    $\lambda_e \lt \lambda_a$

Solution

de-Broglie wavelength is given by $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m q V}}$ For same potential difference $\lambda \propto \frac{1}{\sqrt{m q}}$ $\frac{\lambda_\alpha}{\lambda_e}=\sqrt{\frac{m_e q_e}{m_\alpha q_\alpha}}$ $\because m_\alpha \gg m_e$ $\lambda_e\gt\lambda_\alpha$

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

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