Modern Physics 4 Question 26
27. An $\alpha$-particle and a proton are accelerated from rest by a potential difference of $100 V$. After this, their de-Broglie wavelengths are $\lambda _{\alpha}$ and $\lambda _p$ respectively. The ratio $\frac{\lambda _p}{\lambda _{\alpha}}$, to the nearest integer, is
(2010)
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Solution:
- $\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 q q V m}}$ or $\lambda \propto \frac{1}{\sqrt{q m}}$
$$ \frac{\lambda _p}{\lambda _{\alpha}}=\sqrt{\frac{q _{\alpha}}{q _p} \cdot \frac{m _{\alpha}}{m _p}}=\sqrt{\frac{(2)(4)}{(1)(1)}}=2.828 $$
The nearest integer is 3 .
$\therefore$ Answer is 3 .
