Modern Physics Ques 179
- A particle $A$ of mass $m$ and initial velocity $v$ collides with a particle $B$ of mass $\frac{m}{2}$ which is at rest. The collision is held on, and elastic. The ratio of the de-Broglie wavelengths $\lambda _A$ to $\lambda _B$ after the collision is
(2017 Main)
(a) $\frac{\lambda _A}{\lambda _B}=2$
(b) $\frac{\lambda _A}{\lambda _B}=\frac{2}{3}$
(c) $\frac{\lambda _A}{\lambda _B}=\frac{1}{2}$
(d) $\frac{\lambda _A}{\lambda _B}=\frac{1}{3}$
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Answer:
Correct Answer: 179.(a)
Solution:
Formula:
- For elastic collision,
$ \begin{aligned} p _{\text {before collision }} & =p _{\text {after collision }} \\ m v & =m v _A+\frac{m}{2} v _B \\ 2 v & =2 v _A+v _B \quad …….(i) \end{aligned} $
Now, coefficient of restitution,
$ e=\frac{v _B-v _A}{u _A-v _B} $
Here, $u _B=0$ (Particle at rest) and for elastic collisione $=1$
$ \begin{aligned} & \therefore & 1 & =\frac{v _B-v _A}{v} \\ & \Rightarrow & v & =v _B-v _A \quad …….(ii) \end{aligned} $
From Eq. (i) and Eq. (ii)
$ \begin{gathered} v _A=\frac{v}{3} \text { and } v _B=\frac{4 v}{3} \\ \text { Hence, } \frac{\lambda _A}{\lambda _B}=\frac{(\frac{h}{m V _A})}{\frac{h}{\frac{m}{2} \cdot V _B}}=\frac{V _B}{2 V _A}=\frac{4 / 3}{2 / 3}=2 \end{gathered} $
BETA
