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]
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Correct Answer: Option n
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
\(\frac{1}{2}\) 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