Question: Q. 1. An $\alpha$-particle and a proton are accelerated through the same potential. Find the ratio of their de-Broglie wavelengths.
U] [Delhi 2017]
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Solution:
Ans. de-Broglie wavelength is given by
$$ \lambda=\frac{h}{\sqrt{2 m K}} $$
Mass of a proton $=u$
Charge of proton $=e$
Hence, Kinetic energy of a proton when accelerated through a potential $=\mathrm{eV}$
Putting these values in de-Broglie wavelength formula to calculate wave length of proton
$$ \lambda_{p}=\frac{h}{\sqrt{2 u e V}} $$
Now,
- Mass of an $\alpha$-particle $=4 u$
- Charge of an $\alpha$-particle $=2 e$
Hence, Kinetic energy of an $\alpha$-particle when accelerated through a potential $=2 \mathrm{eV}$
Putting these values in de-Broglie wavelength formula to calculate wave length of $\alpha$-particle
$$ \lambda_{\alpha}=\frac{h}{\sqrt{8 u \times 2 e V}}=\frac{h}{\sqrt{16 u e V}} $$
Hence ratio of $\frac{\lambda_{p}}{\lambda_{\alpha}}=\frac{1}{\sqrt{8}}$
$$ \frac{\lambda_{p}}{\lambda_{\alpha}}=\frac{1}{2 \sqrt{2}} $$
[CBSE Marking Scheme 2017]
