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प्रश्न
In a Geiger-Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 75, when a α-particle of 5 MeV 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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उत्तर
Let r0 be the centre to centre distance between the alpha-particle and nucleus when the α-particle is at its stopping point.
Now`,E_k = 1/(4piepsi_0) ((2e)(ze))/r_0 or r_0 =1/(4piepsi_0 )((2e)(ze))/E_k`
given Ek = 5 × 106eV
= 5 × 106× 1.6 × 10-19V
Z = 75
`r_0 = (9 xx 10^9 xx 75 xx 2(1.6 xx 10^-19)^2)/(5 xx 10^6 xx 1.6 xx 10^-19)`
` =(3456 xx 10 xx^9 xx 10^-38)/(8 xx 10^-13)`
`= 432 xx 10^(+22 -38)`
`=432 xx 10^-16 m`
` =43.2 xx 10^-15 m =43.2 fm`
Since distance of closet approach (r0) is given as
`r_0 = 1/(4piepsi_0) ((2e)(ze))/E_k`
Where Ek is kinetic energy r0 is inversely proportional to Ek.
So when kinetic energy of the α-particle is doubled the distance between closet approach r0 is halved.
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