Question

It is not very difficult to accelerate an electron to a speed that is 99.5% of the speed of light, because it has such a very small mass. what is the ratio of the kinetic energy k to the rest energy mc2 in this case? in the definition of what we mean by kinetic energy (k = e - mc2), you must use the full relativistic formula for e, because v/c is not small compared to 1.

Answer 1

m = m_o / √(1-V^2/c^2 ) = 0.995 c

 K/(m_o c^2 )=((m_o/√(1-v^2/c^2 )) v^2)/(1/2 m_o c^2 )=(m_o/√(1-v^2/c^2 )  (0.995)^2 c^2)/(1/2 m_o c^2 )=(2 x (0.995)^2)/√(1-(0.995c)^2/c^2 )= 198.5 

k/m_oc^2 = 198.5