Electrostatics Question 398
Question: The ratio of momenta of an electron and an a-particle which are accelerated from rest by a potential difference of 100 volt is [UPSEAT 1999]
Options:
A) 1
B) $ \sqrt{\frac{2m _{e}}{{m _{\alpha }}}} $
C) $ \sqrt{\frac{m _{e}}{{m _{\alpha }}}} $
D) $ \sqrt{\frac{m _{e}}{2{m _{\alpha }}}} $
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Answer:
Correct Answer: D
Solution:
Momentum $ p=\sqrt{2mK}; $ where K = kinetic energy = Q.V therefore $ p=\sqrt{2mQV} $
therefore $ p\propto \sqrt{mQ}\Rightarrow \frac{p _{e}}{{p _{\alpha }}}=\sqrt{\frac{m _{e}Q _{e}}{{m _{\alpha }}{Q _{\alpha }}}} $
$ =\sqrt{\frac{m _{e}}{2{m _{\alpha }}}} $
BETA
