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...
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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]
Answer Details
An electron accelerated through a potential difference \(V\) gains kinetic energy equal to the electrical work done on it:
\[ eV = \frac{1}{2}m v^{2} \Rightarrow v = \sqrt{\frac{2eV}{m}} \]Data: \(V = 2.5\,\text{kV} = 2.5\times10^{3}\,\text{V}\); \(e = 1.6\times10^{-19}\,\text{C}\); \(m = 9.1\times10^{-31}\,\text{kg}\).
\[ v = \sqrt{\frac{2(1.6\times10^{-19})(2.5\times10^{3})}{9.1\times10^{-31}}} = \sqrt{\frac{8.0\times10^{-16}}{9.1\times10^{-31}}} \] \[ v = \sqrt{8.79\times10^{14}} = 2.97\times10^{7}\,\text{ms}^{-1} \]The maximum speed of the accelerated electrons is about \(3.0\times10^{7}\,\text{ms}^{-1}\).
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