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


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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

	$\frac{1}{2} {m v}^{2} = \frac{2}{100} m_{0}$ $\frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} \times v^{2} = \frac{1}{25} m_{0}$ $25 v^{2} = \sqrt{1 - \frac{v^{2}}{c^{2}}}$ $25 v^{2} \approx 1 - \frac{1}{2} \frac{v^{2}}{c^{2}}$ $25 v^{2} + \frac{1}{2} \frac{v^{2}}{c^{2}} = 1$ $\frac{50 v^{2} c^{2} + v^{2}}{2 c^{2}} = 1$ $v^{2} (50 c^{2} + 1) = 2 c^{2}$ $v = \sqrt{\frac{2 c^{2}}{1 + 50 c^{2}}} c = s p e e d o f l i g h t . . .$
	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

	$s t u d y r e l a t i v e t h e o r y . . .$
	Commented by Necxx last updated on 03/Aug/18

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