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...
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]
An electron accelerated through a potential difference \(V\) gains kinetic energy equal to the electrical work done on it:
\[ eV = \frac{1}{2}m v^{2} \Rightarrow v = \sqrt{\frac{2eV}{m}} \]