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An $\alpha$-particle and a proton are accelerated from rest by the same potential, then the ratio of their de-Broglie wavelength is
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Verified Answer
The correct answer is:
$1: 2 \sqrt{2}$
de-Broglie's wavelength,
$$
\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m e V}}
$$
The particles are at same potential, so
$$
\frac{\lambda_\alpha}{\lambda_p}=\frac{h / \sqrt{2 m_\alpha e_\alpha V}}{h / \sqrt{2 m_p e_p V}}
$$
Mass of $\alpha$-particle $=4$ times mass of proton.
Charge of $\alpha$-particle $=2$ times charge of proton.
So, $\quad \frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{m_p e_p}{4 m_p 2 e_p}}=\frac{1}{2 \sqrt{2}}$
$$
\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m E}}=\frac{h}{\sqrt{2 m e V}}
$$
The particles are at same potential, so
$$
\frac{\lambda_\alpha}{\lambda_p}=\frac{h / \sqrt{2 m_\alpha e_\alpha V}}{h / \sqrt{2 m_p e_p V}}
$$
Mass of $\alpha$-particle $=4$ times mass of proton.
Charge of $\alpha$-particle $=2$ times charge of proton.
So, $\quad \frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{m_p e_p}{4 m_p 2 e_p}}=\frac{1}{2 \sqrt{2}}$
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