Nuclear Physics And Radioactivity Question 50
Question: The rest energy of an electron is 0.511 MeV. The electron is accelerated from rest to a velocity 0.5 c. The change in its energy will be [MP PET 1996]
Options:
A) 0.026 MeV
B) 0.051 MeV
C) 0.079 MeV
D) 0.105 MeV
Show Answer
Answer:
Correct Answer: C
Solution:
$ \Delta =mc^{2}-m_{0}c^{2}=\frac{m_{0}c^{2}}{\sqrt{1-(v^{2}/c^{2})}}-m_{0}c^{2} $ $ =m_{0}c^{2}( \frac{1}{\sqrt{1-(v^{2}/c^{2})}}-1 )=0.511,( \frac{1}{\sqrt{0.75}}-1 ) $ = 0.079MeV
BETA
