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An $\alpha$-particle and a proton are accelerated from rest by a potential difference of $100 \mathrm{~V}$. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_p$ respectively. The ratio $\frac{\lambda_p}{\lambda_\alpha}$, to the nearest integer, is
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Verified Answer
The correct answer is:
3
$$
\begin{aligned}
\lambda & =\frac{h}{P}=\frac{h}{\sqrt{2 q q V m}} \text { or } \lambda \propto \frac{1}{\sqrt{q m}} \\
\frac{\lambda_p}{\lambda_\alpha} & =\sqrt{\frac{q_\alpha}{q_p} \cdot \frac{m_{\propto}}{m_p}}=\sqrt{\frac{(2)(4)}{(1)(1)}}=2.828
\end{aligned}
$$
The nearest integer is 3 .
$\therefore$ answer is 3 .
\begin{aligned}
\lambda & =\frac{h}{P}=\frac{h}{\sqrt{2 q q V m}} \text { or } \lambda \propto \frac{1}{\sqrt{q m}} \\
\frac{\lambda_p}{\lambda_\alpha} & =\sqrt{\frac{q_\alpha}{q_p} \cdot \frac{m_{\propto}}{m_p}}=\sqrt{\frac{(2)(4)}{(1)(1)}}=2.828
\end{aligned}
$$
The nearest integer is 3 .
$\therefore$ answer is 3 .
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