Question
An $ \alpha $-particle and a proton are accelerated through the same potential difference. Find the ratio of their de-Broglie wavelength.
✓
Answer
$ \lambda=\frac{h}{p}\Rightarrow\lambda=\frac{h}{\sqrt{2mE}} $ or $ \lambda=\frac{h}{\sqrt{2mqV}} $
$\therefore \quad \frac{\lambda_\alpha}{\lambda_p}=\frac{h / \sqrt{2 m_\alpha \cdot q_\alpha V }}{h / \sqrt{2 m_p q_p \cdot V}}=\sqrt{\left(\frac{m_p}{m_\alpha}\right)\left(\frac{q_p}{q_\alpha}\right)}$
$ =\sqrt{(\frac{m_{p}}{4m_{p}})(\frac{q_{p}}{2q_{p}})}=\sqrt{(\frac{1}{8})} $
$\Rightarrow \quad \lambda_\alpha: \lambda_p=1: \sqrt{8}$ or $\lambda_\alpha: \lambda_p=1: 2 \sqrt{2}$
$\therefore \quad \frac{\lambda_\alpha}{\lambda_p}=\frac{h / \sqrt{2 m_\alpha \cdot q_\alpha V }}{h / \sqrt{2 m_p q_p \cdot V}}=\sqrt{\left(\frac{m_p}{m_\alpha}\right)\left(\frac{q_p}{q_\alpha}\right)}$
$ =\sqrt{(\frac{m_{p}}{4m_{p}})(\frac{q_{p}}{2q_{p}})}=\sqrt{(\frac{1}{8})} $
$\Rightarrow \quad \lambda_\alpha: \lambda_p=1: \sqrt{8}$ or $\lambda_\alpha: \lambda_p=1: 2 \sqrt{2}$
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