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A projectile of mass m, charge Z., initial speed v and impact parameter b is scattered by a heavy nucleus of charge Z. Use angular momentum and energy conservation to obtain a formula connecting the minimum distance (s) of the projectile form the nucleus to these parameters. Show that for b = 0, s reduces to the closest distance of approach r_(0). |
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Answer» Solution :Fig. shows the conditions of the problem. Charge on the nucleus = Ze Charge on the projectile = Z.E  At infinity, the angular momentum of projectile about the nucleus = mvb When the projectile is at the minimum DISTANCE s from the nucleus, its velocity vector v. is normal to the radius vector (drawn from the centre of the nucleus). `:.` Angular momentum of the projectile about the nucleus = mv.s From the law of conservation of angular momentum, we have, mvb = mv.s or `v.=(vb)/(s)""_______(i)` When the projectile is infinitely away from the nucleus, it has only KINETIC energy (`because` P.E = 0) given by, `E_(k)=(1)/(2)mv^(2)` When the projectile is at minimum distance s from the nucleus, it has both kinetic and potential energies given by, `E_(k)=(1)/(2)mv.^(2)` and `E._(p)=(1)/(4pi epsilon_(0)).("ZZ".e^(2))/(s)` From the law of conservation of energy, we have, `E_(k)+E_(p)=E_(k)` or `(1)/(2)mv.^(2)+(1)/(4pi epsilon_(0)).("ZZ".e^(2))/(s)=(1)/(2)mv^(2)` PUTTING v. = vb/s from eq. (i), we get, `(1)/(2)m(v^(2)b^(2))/(s^(2))+(1)/(4pi epsilon_(0)).("ZZ".e^(2))/(s)=(1)/(2)mv^(2)` Dividing both sides by `mv^(2)//2`, we get, `(b^(2))/(s^(2))+(1)/(4pi epsilon_(0)).("ZZ".e^(2))/(s(mv^(2)//2))=1` or `s^(2)=b^(2)+(1)/(4pi epsilon_(0)).("ZZ".e^(2)s)/(mv^(2)//2)` This is the required formula. For a head-on collision, b = 0. `:. s^(2)=(1)/(4pi epsilon_(0)).("ZZ".e^(2)s)/(mv^(2)//2):.s=(1)/(4pi epsilon_(0)).("ZZ".e^(2)s)/(mv^(2)//2)` which is the distance of closest approach |
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