KCET · Physics · Dual Nature of Matter
A proton and an alpha particle are accelerated under the same potential difference. The ratio of de-Broglie wavelengths of the proton and the alpha particle is
- A 2
- B \(\sqrt{8}\)
- C \(\frac{1}{\sqrt{8}}\)
- D 1
Answer & Solution
Correct Answer
(B) \(\sqrt{8}\)
Step-by-step Solution
Detailed explanation
Kinetic energy of charged particle, accelerated through a potential difference \(\mathrm{V}\),
\[
\text { i.e., } \mathrm{K}=\mathrm{qV}
\]
\(\therefore\) Momentum of the particle,
\[
\mathrm{p}=\sqrt{2 \mathrm{mK}}=\sqrt{2 \mathrm{mqV}}
\]
de-Broglie wavelength of the particle
\[
\begin{aligned}
\lambda &=\frac{h}{p}=\frac{h}{\sqrt{2 m q v}} \\
\because \quad m_{\alpha} &=m_{p} \text { and } q=2 c \\
\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_{\alpha}} &=\sqrt{\frac{m_{2} q_{2}}{m_{1} q_{1}}}=\sqrt{\frac{4 \times 2}{1 \times 1}}=\sqrt{8}
\end{aligned}
\]
\[
\text { i.e., } \mathrm{K}=\mathrm{qV}
\]
\(\therefore\) Momentum of the particle,
\[
\mathrm{p}=\sqrt{2 \mathrm{mK}}=\sqrt{2 \mathrm{mqV}}
\]
de-Broglie wavelength of the particle
\[
\begin{aligned}
\lambda &=\frac{h}{p}=\frac{h}{\sqrt{2 m q v}} \\
\because \quad m_{\alpha} &=m_{p} \text { and } q=2 c \\
\therefore \quad \frac{\lambda_{\mathrm{p}}}{\lambda_{\alpha}} &=\sqrt{\frac{m_{2} q_{2}}{m_{1} q_{1}}}=\sqrt{\frac{4 \times 2}{1 \times 1}}=\sqrt{8}
\end{aligned}
\]
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