An elementary particle of mass m and charge +e is projected with velocity v at a much more massive fixed particle of charge Ze, where Z > 0. What is the closest possible approach of the incident particle?
\( \frac{Ze^2}{2\pi\varepsilon_0 m v^2} \)
\( \frac{Ze}{4\pi\varepsilon_0 m v^2} \)
\( \frac{Ze^2}{8\pi\varepsilon_0 m v^2} \)
\( \frac{Ze}{8\pi\varepsilon_0 m v^2} \)
Solution:
At the distance of closest approach \( r_0 \), the kinetic energy is completely converted into electrostatic potential energy: \( \frac{1}{2} m v^2 = \frac{1}{4 \pi \varepsilon_0} \frac{Ze^2}{r_0} \). Rearranging gives \( r_0 = \frac{Ze^2}{2\pi\varepsilon_0 m v^2} \).
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