Question

At what speed the mass of electron is double ?

Answer 1

The relativistic mass of an electron is given by,

m_{relative} = \frac{ m_{o} }{ \sqrt{1 - v^{2} } / c^{2} }mrelative=1−v2/c2mo 

To find when the speed at which the mass of an electron doubles, we equate

m_{relative} = 2 m_{o}mrelative=2mo 

So we get,

\frac{1}{ \sqrt{1 - v^{2}/ c^{2} } } = 21−v2/c21=2 

\sqrt{1 - v^{2}/ c^{2} } = 1/21−v2/c2=1/2 

1 - v^{2}/ c^{2} = 1/41−v2/c2=1/4 

v^{2}/ c^{2} = 3/4v2/c2=3/4 

v = 3/4 c [c = speed of light]

v = 3/4 c