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Q.

Rest mass energy of an electron is 0.51 MeV. If this electron is moving with a velocity 0.8 c (where c is velocity of light in vacuum), then kinetic energy of the electron should be.

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a

0.34 MeV

b

0.46 MeV

c

0.28 MeV

d

0.39 MeV

answer is B.

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Detailed Solution

Given m0c2 = 0.51 MeV and v = 0.8 c
K.E. of the electron = mc2– m0c2
But  m\;=\;\frac{m_{0}}{\sqrt{1-\frac{V^{2}}{C^{2}}}}\;=\;\frac{m_{0}}{\sqrt{1-\left ( \frac{0.8c}{c} \right )^{2}}}\;=\;\frac{m_{0}}{\sqrt{0.36}}\;=\;\frac{m_{0}}{0.6}
Now,m{c^2} = \frac{{0.51}}{{0.6}}  MeV = 0.85 MeV
\therefore K.E. = (0.85 – 0.51)MeV = 0.34 MeV
 

 

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