Dual Nature Of Light Question 86
Question: A particle of mass M at rest decays into two particles of masses m1 and m2, having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles, $ {\lambda_{1}}/{\lambda_{2}} $ is [IIT-JEE 1999; KCET 2003]
Options:
A) $ m_{1}/m_{2} $
B) $ m_{2}/m_{1} $
C) 1.0
D) $ \sqrt{m_{2}}/\sqrt{m_{1}} $
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Answer:
Correct Answer: C
Solution:
By law of conservation of momentum $ 0=m_{1}\overrightarrow{v_{1}}+m_{2}\overrightarrow{v_{2}}\Rightarrow m_{1}\overrightarrow{v_{1}}=-m_{2}\overrightarrow{v_{2}} $ ? ve sign indicates that both he particles are moving in opposite direction. Now de-Broglie wavelengths $ {\lambda_{1}}=\frac{h}{m_{1}v_{1}} $ and $ {\lambda_{2}}=\frac{h}{m_{2}v_{2}} $ ;
$ \therefore \ \frac{{\lambda_{1}}}{{\lambda_{2}}}=\frac{m_{2}v_{2}}{m_{1}v_{1}}=1 $
BETA
