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A particle of charge $q$, mass $m$ and energy $E$ has de-Broglie wavelength $\lambda$. For a particle of charge $2 q$, mass $2 m$ and energy $2 E$, the de-Broglie wavelength is
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Verified Answer
The correct answer is:
$\frac{\lambda}{2}$
The de-Broglie wavelength $\lambda$ of charged particle of charge $q$ mass $m$ and energy $E$ is given by
$$
\lambda=\frac{h}{\sqrt{2 m E}}
$$
For another particle mass $m^{\prime}=2 \mathrm{~m}$
charge, $q^{\prime}=2 q$
and energy, $B^{\prime}=2 B$
de-Broglie wavelength,
$$
\begin{aligned}
\lambda^{\prime} & =\frac{h}{\sqrt{2 m^{\prime} B^{\prime}}} \\
& =\frac{h}{\sqrt{2 \times 2 m \times 2 E}} \\
& =\frac{1}{2} \cdot \frac{h}{\sqrt{2 m E}} \lambda^{\prime}=\frac{\lambda}{2}
\end{aligned}
$$
$$
\lambda=\frac{h}{\sqrt{2 m E}}
$$
For another particle mass $m^{\prime}=2 \mathrm{~m}$
charge, $q^{\prime}=2 q$
and energy, $B^{\prime}=2 B$
de-Broglie wavelength,
$$
\begin{aligned}
\lambda^{\prime} & =\frac{h}{\sqrt{2 m^{\prime} B^{\prime}}} \\
& =\frac{h}{\sqrt{2 \times 2 m \times 2 E}} \\
& =\frac{1}{2} \cdot \frac{h}{\sqrt{2 m E}} \lambda^{\prime}=\frac{\lambda}{2}
\end{aligned}
$$
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