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11.Dual Nature of Radiation and matter
hard
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. ........... $MeV$
A
$0.28 $
B
$0.34$
C
$0.39$
D
$0.46$
Solution
(b) Given $m_0c^2$ $= 0.51 \ MeV$ and $v = 0.8\ c$
$K.E$. of the electron = $mc^2$-$m_0c^2$
But $m = \frac{{{m_0}}}{{\sqrt {1 – \frac{{{v^2}}}{{{c^2}}}} }} = \frac{{{m_0}}}{{\sqrt {1 – {{\left( {\frac{{0.8\,c}}{c}} \right)}^2}} }} = \frac{{{m_0}}}{{\sqrt {0.36} }} = \frac{{{m_0}}}{{0.6}}$
Now, $m{c^2} = \frac{{0.51}}{{0.6}}$ $MeV = 0.85\ MeV $
$\therefore K.E. = (0.85 -0.51)\ MeV = 0.34\ MeV.$
Standard 12
Physics
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