for scattering by an inverse square law field (such as that produced by a charged nucleus in Rutherford's model), the relation between impact parameter b and the scattering angle `theta` is given by
`b=(Ze^2 cot theta//2)/(4pi in_0(1/2mv^2))`
(a) What is the scattering angle for b=0?
(b) for given impact parameter, b, does the angle of deflection increase or decrease with increase in energy?
(c) What is the impact parameter at which the scattering angle is `90^@` for Z=79 and initial energy=10MeV?
(d) Why is it that the mass of the nucleus does not enter the formula above, but its charge does?
(e) for a given energy of the projectile, does the scattering angle increase or decrease with decrease in impact parameter?

Explanation

(a) The given relation in Rutherford atom model is
`b=(Ze^2 cot theta//2)/(4pi in_0 (1/2mv^2))`
When b=0, `cot theta//2=0`
(`because` all other quantities are finite)
`:. (theta)/(2) = 90^(@)` or `theta=180^(@)`
i.e, Alpha particle is scatterd through `180^@`, when impact parameter is zero
(b) When b is given (i.e. constant)
`(Ze^2 cot theta//2)/(4pi in_0 (1/2mv^2))`=constant
`:. cot.(theta)/(2)` increase with increase in energy
`E(=1/2mv^2)`
Therefore, `theta` decrease with increase of energy.
(c) Here, `theta=90^@, Z=79,`
`E=1/2mv^2=10MeV=1.6xx10^(-12)"joule"`.
`1/(4pi in_0)=9xx10^9 Nm^2C^-2,b=?`
As `b=(Ze^2 cot theta//2)/(4pi in_0 (1/2mv^2))`
`:. b=(9xx10^9xx79(1.6xx10^(-19))^2cot 45^@)/(1.6xx10^(-12))`
`=9xx79xx1.6xx10^(-17)m`
`=1.137xx10^(-14)m`
(d) Scattering takes place due to charge on the nucleus. If `Z=0`,`theta=0` form the given formula. Mass of the nucleus does not enter the formula, as we are ignoring the recoil of the nucleus. A small correction term will appear in the formula involving mass of the nucleus, if we do not ignore the recoil of the nucleus.
(e) for a given energy `(E=1/2mv^2`, fixed), the decrease in impact parameter (b) implies decrease in `(cot theta//2)`. Hence `theta` increase with decrease in impact parameter.