Modern Physics Ques 158
- A particle of mass $M$ at rest decays into two particles of masses $m _1$ and $m _2$ having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles $\lambda _1 / \lambda _2$ is
$(1999,2 M)$
(a) $m _1 / m _2$
(b) $m _2 / m _1$
(c) $1$
(d) $\sqrt{m _2} / \sqrt{m _1}$
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Answer:
Correct Answer: 158.(c)
Solution:
Formula:
- From law of conservation of momentum,
$ p _1=p _2 \quad \text { (in opposite directions) } $
Now de-Broglie wavelength is given by
$ \lambda=\frac{h}{p}, \text { where } h=\text { Planck’s constant } $
Since magnitude of momentum $(p)$ of both the particles is equal, therefore $\lambda _1=\lambda _2$ or $\lambda _1 / \lambda _2=1$
BETA
