Find the speed which the mass of an electron is double of its rest mass-

Question

Toppr Admin · Accepted Answer

The mass of an electron at speed v is-xA0-m-m0-x221A-1-x2212-v2-c2where m0 is its rest mass- If m-2m0-2-1-x221A-1-x2212-v2-c2or- -xA0- -xA0-1-x2212-v2c2-14or- v-x221A-32c-2-598-xD7-108ms-x2212-1-

Find the speed at which the mass of an electron is double of its rest mass.

The mass of an electron at speed $v$ is
$m = \frac{m_{0}}{\sqrt{1 - v^{2} / c^{2}}}$
where $m_{0}$ is its rest mass. If $m = 2 m_{0}$ ,
$2 = \frac{1}{\sqrt{1 - v^{2} / c^{2}}}$
or, $1 - \frac{v^{2}}{c^{2}} = \frac{1}{4}$
or, $v = \frac{\sqrt{3}}{2} c = 2.598 \times 10^{8} m s^{- 1}$ .

Find the speed at which the mass of an electron is double of its rest mass.

The mass of an electron at speed v is m=m0√1−v2/c2where m0 is its rest mass. If m=2m0,2=1√1−v2/c2or, 1−v2c2=14or, v=√32c=2.598×108ms−1.

The mass of an electron at speed $v$ is
$m = \frac{m_{0}}{\sqrt{1 - v^{2} / c^{2}}}$
where $m_{0}$ is its rest mass. If $m = 2 m_{0}$ ,
$2 = \frac{1}{\sqrt{1 - v^{2} / c^{2}}}$
or, $1 - \frac{v^{2}}{c^{2}} = \frac{1}{4}$
or, $v = \frac{\sqrt{3}}{2} c = 2.598 \times 10^{8} m s^{- 1}$ .