A photon with initial frequency 10^(11)Hz scatters off an electron at rest. Its final frequency is 0.9 xx 10^(11)Hz. Calculate the speed of the scattered electron (h = 6.63 xx 10^(-34) Js, m_(e) = 9.1 xx 10^(-31) kg).

A photon with initial frequency 10^(11)Hz scatters off an electron at

1.	A photon with initial frequency 10^(11)Hz scatters off an electron at rest. Its final frequency is 0.9 xx 10^(11)Hz. Calculate the speed of the scattered electron (h = 6.63 xx 10^(-34) Js, m_(e) = 9.1 xx 10^(-31) kg).
Answer» SOLUTION :INCIDENT energy = Scattered energy + K.E. of the scattered electron `hv_(1) = hv_(2) + (1)/(2) MV^(2)` or `(1)/(2) mv^(2) = h (v_(1) - v_(2))` or `V^(2) = (2h(v_(1) - v_(2)))/(m) = (2 xx 6.63 xx 10^(-34) kg m^(2) s^(-1) (10^(11) - 0.9 xx 10^(11)) s^(-1))/(9.1 xx 10^(-31) kg)` `= (2 xx 6.63 xx 10^(-34) xx 0.1 xx 10^(11))/(9.1 xx 10^(-31)) m^(2) s^(-2) = 14571428 m^(2) s^(-2)` or `v = sqrt(14571428) ms^(-1) = 3.8 xx 10^(3) ms^(-1)`