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A proton and an alpha particle are subjected to same potential difference $V$. Their de-Broglie wavelengths $\lambda_{p}, \lambda_{\alpha}$ will be in the ratio
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Verified Answer
The correct answer is:
$2 \sqrt{2}: 1$
T
he de-Broglie wavelength of a particle is given by
$\lambda=\frac{h}{m v}=\frac{h}{p}$
The momentum of proton,
$p_{p}=\sqrt{2 m E}=\sqrt{2 m e V}$
Similarly, for $\alpha$ - particle,
$\begin{aligned}
p_{\alpha} &-\sqrt{2 \times 4 m \times 2 e \times V} \\
&=\sqrt{16 m e V} \\
\therefore \quad \frac{\lambda_{p}}{\lambda_{\alpha}} &=\frac{p_{\alpha}}{p_{p}}=\frac{\sqrt{16 m e V}}{\sqrt{2 m e V}}=\sqrt{8} \\
&=2 \sqrt{2}: 1
\end{aligned}$
he de-Broglie wavelength of a particle is given by
$\lambda=\frac{h}{m v}=\frac{h}{p}$
The momentum of proton,
$p_{p}=\sqrt{2 m E}=\sqrt{2 m e V}$
Similarly, for $\alpha$ - particle,
$\begin{aligned}
p_{\alpha} &-\sqrt{2 \times 4 m \times 2 e \times V} \\
&=\sqrt{16 m e V} \\
\therefore \quad \frac{\lambda_{p}}{\lambda_{\alpha}} &=\frac{p_{\alpha}}{p_{p}}=\frac{\sqrt{16 m e V}}{\sqrt{2 m e V}}=\sqrt{8} \\
&=2 \sqrt{2}: 1
\end{aligned}$
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