Question on relativity Determine the speed at which the kine

Question on relativity

Determine the speed at which the kinetic energy of an electron is equal to twice its rest energy. 0.45c 0.63c 0.87c 0.94c 0.99c

Given

KE = ymc^2 - mc^2 = 2*mc^2

which means

y = 3

y = 1/sqrt(1 - v^2/c^2) = 3

1 - v^2/c^2 = 1/9

v^2/c^2 = 8/9

v = c*sqrt(8/9) = 0.942*c