Question
A proton and an $\alpha$-particle are accelerated from rest by $2\,V$ and $4\,V$ potentials, respectively. The ratio of their de-Broglie wavelength is:
✓
Answer
$\lambda=\frac{ h }{ mv }=\frac{ h }{\sqrt{2 mK }}=\frac{ h }{\sqrt{2 mq \Delta V }}$
$\frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{m_p V_p q_p}{ m _\alpha V_\alpha q_\alpha}}$
$\Rightarrow \frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{1 \times 2 \times 1}{4 \times 4 \times 2}}=\frac{1}{4}$
$\Rightarrow \lambda_p: \lambda_\alpha=4: 1$
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