Q37 / 50 · Manipal (MET) 2023 · Physics

A proton accelerated through a potential difference of 100 V , has de-Broglie wavelength $\lambda_0$. The de-Broglie wavelength of an $\alpha$-particle, accelerated through 800 V is

  • A
    $\frac{\lambda_0}{\sqrt{2}}$
  • B
    $\frac{\lambda_0}{2}$
  • C
    $\frac{\lambda_0}{4}$
  • Correct answer
    D
    $\frac{\lambda_0}{8}$

Solution

As, $\lambda=\frac{h}{\sqrt{2 q V m}} \propto \frac{1}{\sqrt{q V m}}$ $\begin{array}{ll}\therefore & \frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{q_p V_p m_p}{q_\alpha V_\alpha m_\alpha}}=\sqrt{\frac{1 \times 100 \times 1}{2 \times 800 \times 4}}=\frac{1}{8} \\ \therefore & \lambda_\alpha=\frac{\lambda_p}{8}=\frac{\lambda_0}{8}\end{array}$

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

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