Question 29: Dual Nature of Matter and Radiation

Question

A proton accelerated through a potential V has de-Broglie wavelength $\lambda$. Then, the de-Broglie wavelength of an $\alpha$-particle, when accelerated through the same potential V is

Accepted Answer

When a charged particle of charge $q$ is accelerated through a potential $V$, then de-Broglie wavelength is given as
$\begin{aligned}& \lambda=\frac{h}{\sqrt{2 m q V}} \
& \Rightarrow \quad \lambda \propto \frac{1}{\sqrt{m q}} \
& \Rightarrow \quad \frac{\lambda_\alpha}{\lambda_p}=\sqrt{\frac{m_p}{m_\alpha}} \cdot \sqrt{\frac{q_p}{q_\alpha}} \
& \Rightarrow \quad \frac{\lambda_\alpha}{\lambda}=\sqrt{\frac{m_p}{4 m_p}} \sqrt{\frac{q_p}{2 q_p}} \
& {\left[\begin{array}{ll}
\because m_\alpha=4 m_p \
	ext { and } q_\alpha=2 q_p
\end{array}ight] } \
& \Rightarrow \quad \frac{\lambda_\alpha}{\lambda}=\frac{1}{\sqrt{8}} \
& \Rightarrow \quad \lambda_\alpha=\frac{\lambda}{2 \sqrt{2}}
\end{aligned}
$

Answer

$\frac{\lambda}{2}$

Answer

$\frac{\lambda}{\sqrt2}$

Answer

$\frac{\lambda}{2\sqrt2}$

Answer

$\frac{\lambda}{8}$

Dual Nature of Matter and Radiation

A proton accelerated through a potential V has de-Broglie wavelength \lambda. Then, the de-Broglie wavelength of an \alpha-particle, when accelerated through the same potential V is

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