Q146 / 180 · KCET 2018 · Physics
A proton and an \( \alpha \) particle are accelerated through the same potential difference \( V \). The ratio of
their de-Broglie's wavelength is
their de-Broglie's wavelength is
- A\( \sqrt{2} \)
- Correct answerB\( 2 \sqrt{2} \)
- C\( \sqrt{3} \)
- D\( 2 \sqrt{3} \)
Solution
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
\[ \frac{\lambda_{p}}{\lambda_{a}}=\frac{\sqrt{m_{a} q_{a}}}{\sqrt{m_{p} q_{p}}}=\frac{\sqrt{4 m \times 2 e}}{\sqrt{m \times e}}=\sqrt{8}=2 \sqrt{2} \]
Hence, the correct answer is option: 'B'.