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Question
In a Geiger-Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when a α-particle of 8Mev energy impinges on it before it comes momentarily to rest and reverses its direction.
How will the distance of closest approach be affected when the kinetic energy of the α-particle is doubles?
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Solution
Let r0 be the centre to centre distance between the alpha-particle and nucleus when the ∝-particle is at its stopping point, Given Z = 80, Ek = 8MeV
Now, `E_k = 1/(4piepsi_0) ((Ze)(2e))/r_0`
`so r_0 = 1/(4piepsi_0) ((2e)(Ze))/E_k`
`= (9 xx 10^9 xx 80 xx 2 (1.6 xx 10^-19))^2/(8 xx 10^6 xx 1.6 xx 10^-19)`
`= (18 xx 80 xx 10^9 xx 1.6 xx 10^-19)/(8 xx 10^6)`
`= 28.8 fm`
Since `r_0 ∝ 1/E_k`
So when kinetic energy is doubled the distance of closest approach r0 is halved.
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