Modern Physics Ques 176
- A particle $A$ of mass ’ $m$ ’ and charge ’ $q$ ’ is accelerated by a potential difference of $50$ $ V$. Another particle $B$ of mass ’ $4 $ $m$ ’ and charge ’ $q$ ’ is accelerated by a potential difference of $2500$ $ V$. The ratio of de-Broglie wavelengths $\frac{\lambda _A}{\lambda _B}$ is close to
(a) $4.47$
(b) $10.00$
(c) $0.07$
(d) $14.14$
(Main 2019, 12 Jan I)
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Answer:
Correct Answer: 176.(d)
Solution:
Formula:
- de-Broglie wavelength associated with a moving charged particle of charge $q$ is
$ \lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m q V}} $
where, $V=$ accelerating potential.
Ratio of de-Broglie wavelength for particle $A$ and $B$ is,
$ \frac{\lambda _A}{\lambda _B}=\frac{\sqrt{m _B q _B V _B}}{\sqrt{m _A q _A V _A}}=\sqrt{\frac{m _B}{m _A}} \cdot \sqrt{\frac{q _B}{q _A}} \cdot \sqrt{\frac{V _B}{V _A}} $
Substituting the given values, we get.
$ \begin{aligned} & =\sqrt{\frac{4 m}{m}} \cdot \sqrt{\frac{q}{q}} \cdot \sqrt{\frac{2500}{50}} \\ & =2 \times 1 \times 5 \times 1.414=14.14 \end{aligned} $
BETA
