CBSE 2026 · Region 1 · Set 2 · Q18 · 2 marks
Find the ratio $\displaystyle \left(\frac{\lambda_{\mathrm{a}}}{\lambda_{\mathrm{p}}}\right)$ of de Broglie wavelength $\displaystyle \lambda_{\mathrm{a}}$ associated with an alpha particle to de Broglie wavelength $\displaystyle \lambda_{\mathrm{p}}$ associated with a proton if both are moving with the (a) same velocity (b) same kinetic energy.
Marking-scheme solution
de-Broglie wavelength $\displaystyle \lambda=\dfrac{h}{mv}$
(a)
For same velocity $\displaystyle \lambda \alpha \dfrac{1}{m}$
$\displaystyle \dfrac{\lambda_{a}}{\lambda_{p}}=\dfrac{m_{p}}{m_{a}}$
$\displaystyle \because$ mass of alpha is four times of mass of proton. Hence $\displaystyle \dfrac{\lambda_{a}}{\lambda_{p}}=\dfrac{m_{p}}{4 m_{p}}$
$\displaystyle \dfrac{\lambda_{a}}{\lambda_{p}}=\dfrac{1}{4}$
(b)
For same kinetic energy $\displaystyle \lambda=\dfrac{h}{\sqrt{2 m K}}$
$\displaystyle \lambda \alpha \dfrac{1}{\sqrt{m}}$
$\displaystyle \dfrac{\lambda_{a}}{\lambda_{p}}=\dfrac{1}{2}$
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.