Determine the speed at which the kinetic energy of a particle will deviate by 1 % from its nonrelativistic value 1 2 m₀ v^2 .

Step-by-step solution

The relativistic kinetic energy is given by K = m₀ c^2 ( - 1) = m₀ c^2 ( 1 1 - v^2 c^2 - 1 ) . Using the binomial expansion for v c , we get: K = m₀ c^2 ( 1 + 1 2 v^2 c^2 + 3 8 v^4 c^4 + - 1 ) K 1 2 m₀ v^2 + 3 8 m₀ v^4 c^2 The nonrelativistic kinetic energy is K_ nr = 1 2 m₀ v^2 . The fractional deviation is given by: K - K_ nr K_ nr = 3 8 m₀ v^4 c^2 1 2 m₀ v^2 = 3 4 v^2 c^2 Given that the deviation is 1 % = 0.01 : 3 4 v^2 c^2 = 0.01 v^2 = 0.04 3 c^2 v = 0.2 3 c Substituting c = 3 10^8 m/s : v = 0.2 3 3 10^8 = 0.2