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Question
Define the distance of closest approach. An α-particle of kinetic energy 'K' is bombarded on a thin gold foil. The distance of the closest approach is 'r'. What will be the distance of closest approach for an α-particle of double the kinetic energy?
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Solution
Distance of closest approach
When the alpha particle moves towards the nucleus, it's kinetic energy keeps on decreasing because kinetic energy is being consumed in doing work against the force of repulsion of the nucleus. So the distance from the nucleus at which velocity of the alpha particle becomes zero is known as the distance of closest approach.
Distance of closest approach is given as
`(2Z_e^2)/(4π∈_oK)`
or distance of closest approach prop 1/K
or `r_1/r_2=K_2/K_1`
or `r2 = K_1/K_2×r_1`
`= K/(2K)xxr`
`= r/2`
Hence, the distance of closest approach would be r/2.
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