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Determine the accelration of a relativistic electron moving along a unifrom electric field of strength `E` at the moment when its kinetic energy becomes equal to `T`. |
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Answer» As before, `T = e E x` Now in linear motion, `(d)/(dt) (m_(0) v_(x))/(sqrt(1 - v^(2)//c^(2))) = (m_(0) w)/(sqrt(1 - v^(2)//c^(2))) + (m_(0) w)/((1 - v^(2)//c^(2))^(3//2)) = (v)/(c^(2)) w` `= (m_(0))/((1 - v^(2)//c^(2))^(3//2))w = ((T + m_(0) c^(2))^(3))/(m_(0)^(2) c^(6)) w = e E`, So, `w = (e E m_(0)^(2) c^(6))/((T + m_(0) c^(2))^(3)) = (e E)/(m_(0)) (1 + (T)/(m_(0) c^(2)))^(-3)` |
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