Calculate-the-speed-of-an-electron-whose-kinetic-energy-is-equal-to-2-of-its-rest-mass-

Question Number 41148 by Necxx last updated on 02/Aug/18

C a l c u l a t e t h e s p e e d o f a n e l e c t r o n

w h o s e k i n e t i c e n e r g y i s e q u a l t o

2 % o f i t s r e s t m a s s .

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Aug/18

\frac{1}{2} {m v}^{2} = \frac{2}{100} m_{0}

\frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} \times v^{2} = \frac{1}{25} m_{0}

25 v^{2} = \sqrt{1 - \frac{v^{2}}{c^{2}}}

25 v^{2} \approx 1 - \frac{1}{2} \frac{v^{2}}{c^{2}}

25 v^{2} + \frac{1}{2} \frac{v^{2}}{c^{2}} = 1

\frac{50 v^{2} c^{2} + v^{2}}{2 c^{2}} = 1

v^{2} (50 c^{2} + 1) = 2 c^{2}

v = \sqrt{\frac{2 c^{2}}{1 + 50 c^{2}}} c = s p e e d o f l i g h t \dots

Commented by Necxx last updated on 03/Aug/18

T h a n k y o u s o m u c h s i r T a n m a y . .

p l e a s e i f I m a y a s k h o w d i d y o u

c o m e a b o u t

{m v}^{2} = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} ? ?

Commented by tanmay.chaudhury50@gmail.com last updated on 03/Aug/18

s t u d y r e l a t i v e t h e o r y \dots

Commented by Necxx last updated on 03/Aug/18

o k s i r \dots . . T h a n k y o u

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