The accelerating potential in a cathode ray oscilloscope is 2.5 kV. Calculate the maximum speed of the accelerated electrons. [ e = 1.6 x 10(^{-19}) C; Me = 9.1 x 10(^{-31}) kg]
Explanation
(frac{1}{2}) MeV(^2) = eV
(V^2 = frac{2eV}{Me} = frac{2 times 1.6 times 10^{-19} times 2500}{9.1 times 10^{-31}})
= 8.79 x 10(^{-14})
V = (sqrt{8.79 times 10^{14}})
= 2.965 x 10(^7) m/s