CBSE 2024 · Region 4 · Set 3 · Q18 · 2 marks
The ratio of de Broglie wavelengths of a proton and a deuteron accelerated by potential $\displaystyle V_p$ and $\displaystyle V_d$ respectively, $\displaystyle \left(\frac{\lambda_p}{\lambda_d}\right)$ is $\displaystyle \frac{1}{2}$. Find $\displaystyle \frac{V_p}{V_d}$.
Marking-scheme solution
De Broglie wavelength of proton:
$\displaystyle \lambda_{p}=\frac{h}{\sqrt{2 m e V_{p}}}$
De Broglie wavelength of deuteron:
$\displaystyle \lambda_{d}=\frac{h}{\sqrt{2(2 m) e V_{d}}}$
$\displaystyle \frac{\lambda_{p}}{\lambda_{d}}=\frac{1}{2}=\sqrt{\frac{2(2 m) e V_{d}}{2 m e V_{p}}}$
On solving:
$\displaystyle \frac{V_{p}}{V_{d}}=8$
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.