Dual Nature Of Light Question 361
Question: A particle A of mass m and initial velocity v m collides with a particle B of mass $ \frac{m}{2} $ which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths $ {\lambda_{A}} $ to $ {\lambda_{B}} $ after the collision is
Options:
A) $ \frac{2}{3} $
B) $ \frac{1}{2} $
C) $ \frac{1}{3} $
D) $ 2 $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] From question, $ m_{A}=M;m=\frac{m}{2} $ $ u_{A}=V\text{ }u_{B}=0 $ Let after collision velocity of $ A=V_{1} $ and velocity of $ B=V_{2} $ Applying law of conservation of momentum, $ mu=mv_{1}+( \frac{m}{2} )v_{2}\text{ or, 2}\mu \text{=2}v_{1}\text{ +}{{v}{2}} $ ?.(i) By law of collision $ e=\frac{v{2}-v_{1}}{u-1}\text{ or }u=v_{2}-v_{1} $ ?(ii) $ [ \because \text{collision is elastic, }e=1 ] $ using eqns (i) and (11) $ v_{1}=\frac{1}{3}\mu \text{ and }v_{2}=\frac{4}{3}u $ De-Broglie wavelength $ \lambda =\frac{h}{p} $
$ \therefore \frac{{\lambda_{A}}}{{\lambda_{B}}}=\frac{P_{B}}{P_{A}}=\frac{\frac{m}{2}\times \frac{4}{3}u}{m\times \frac{1}{3}}=2 $
BETA
