Q154 / 160 · TS EAMCET 2024 (11 May Shift 1) · Physics
The de Broglie wavelength of a proton is twice the de Broglie wavelength of an alpha particle. The ratio of the kinetic energies of the proton and the alpha particle is
- Correct answerA$1: 1$
- B$1: 4$
- C$1: 2$
- D$1: 8$
Solution
The de-Broglie wavelength of a particle is
$\begin{aligned}
& \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}} \Rightarrow \lambda \propto \frac{1}{\sqrt{\mathrm{mk}}} \\
& \therefore \frac{\lambda_{\mathrm{p}}}{\lambda_\alpha}=\sqrt{\frac{\mathrm{m}_\alpha \mathrm{k}_\alpha}{\mathrm{m}_{\mathrm{p}} \mathrm{k}_{\mathrm{p}}}} \Rightarrow \frac{2 \lambda_\alpha}{\lambda_{\mathrm{p}}}=\sqrt{\left(\frac{4}{1}\right) \frac{\mathrm{k}_\alpha}{\mathrm{k}_{\mathrm{p}}}} \\
& \therefore \frac{\mathrm{k}_{\mathrm{p}}}{\mathrm{k}_\alpha}=1: 1
\end{aligned}$
Hence, the correct answer is option: 'A'.