KCET · Physics · Dual Nature of Matter
A proton and an \( \alpha \) particle are accelerated through the same potential difference \( V \). The ratio of
their de-Broglie's wavelength is
- A \( \sqrt{2} \)
- B \( 2 \sqrt{2} \)
- C \( \sqrt{3} \)
- D \( 2 \sqrt{3} \)
Answer & Solution
Correct Answer
(B) \( 2 \sqrt{2} \)
Step-by-step Solution
Detailed explanation
The de Broglie wavelength is defined as
\[
\lambda=\frac{h}{\sqrt{2 m q V}}
\]
Now, for proton \( \lambda_{p}=\frac{h}{\sqrt{2 m_{p} q V}} \)
For \( \alpha- \) particle, \( \lambda_{\alpha}=\frac{h}{\sqrt{2 m_{a} q_{a} V}} \)
Since
\[
m_{p}=4 m, q_{p}=e ; m_{\alpha}=4 m, q_{\alpha}=2 e
\]
Ratio of their de Broglie wavelengths is
\[
\frac{\lambda_{p}}{\lambda_{a}}=\frac{\sqrt{m_{a} q_{a}}}{\sqrt{m_{p} q_{p}}}=\frac{\sqrt{4 m \times 2 e}}{\sqrt{m \times e}}=\sqrt{8}=2 \sqrt{2}
\]
\[
\lambda=\frac{h}{\sqrt{2 m q V}}
\]
Now, for proton \( \lambda_{p}=\frac{h}{\sqrt{2 m_{p} q V}} \)
For \( \alpha- \) particle, \( \lambda_{\alpha}=\frac{h}{\sqrt{2 m_{a} q_{a} V}} \)
Since
\[
m_{p}=4 m, q_{p}=e ; m_{\alpha}=4 m, q_{\alpha}=2 e
\]
Ratio of their de Broglie wavelengths is
\[
\frac{\lambda_{p}}{\lambda_{a}}=\frac{\sqrt{m_{a} q_{a}}}{\sqrt{m_{p} q_{p}}}=\frac{\sqrt{4 m \times 2 e}}{\sqrt{m \times e}}=\sqrt{8}=2 \sqrt{2}
\]
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