Dual Nature Of Light Question 345
Question: A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has greater value of de-Broglie wavelength associated with it, and less momentum?
Options:
A) Proton
B) Deutron
C) Both have equal values
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] de-Broglie wavelength of a charged particle is given by, $ \lambda \propto \frac{1}{\sqrt{mq}} $ If $ m_{p} $ and e are mass and charge of a proton respectively, and, $ m_{D} $ and e are mass and charge of a deuteron respectively, then $ \frac{{\lambda_{p}}}{{\lambda_{D}}}=\sqrt{\frac{m_{D}q_{D}}{m_{p}q_{p}}}=\sqrt{\frac{( 2m_{p} )( e )}{( m_{p} )( e )}}=\sqrt{2} $ $ {\lambda_{p}}=\sqrt{2}{\lambda_{D}} $ Momentum, is given by, $ P=\frac{h}{\lambda }\text{ or, p}\propto \frac{1}{\lambda } $ Where, h = plank’s constant Since the wavelength of a proton is more than that of deuteron thus, the momentum of a proton is lesser than that of deuteron. Hence, the momentum of proton is less.
BETA
