MCQ
The rest energy of an electron is
  • $510\, KeV$
  • B
    $931\, KeV$
  • C
    $510 \,MeV$
  • D
    $931\, MeV$

Answer

Correct option: A.
$510\, KeV$
a
(a) Rest energy of an electron ${r_n} \propto {n^2}$

Here ${m_e} = 9.1 \times {10^{ - 31}}kg$ and $c =$ velocity of light

$\therefore $Rest energy $ = 9.1 \times {10^{ - 31}} \times {(3 \times {10^8})^2}joule$

$ = \frac{{9.1 \times {{10}^{ - 31}} \times {{(3 \times {{10}^8})}^2}}}{{1.6 \times {{10}^{ - 19}}}}eV = 510\;keV$

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