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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 from 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 nuclesu ). `therefore` 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)/(4 pi 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)/(4 pi 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)/(4 pi epsilon_(0)) .("ZZ".e^(2))/(s(mv^(2)/2)) = " or " s^(2) = b^(2) + (1)/(4 pi epsilon_(0)) .("ZZ".e^(2)s)/(mv^(2)//2)` This is the required formula. For a head- on COLLISION , b = 0. `therefore s^(2) = (1)/(4 pi epsilon_(0)) .("ZZ".e^(2)s)/(mv^(2)//2) " or " therefore s = (1)/(4 pi epsilon_(0)) .("ZZ".e^(2)s)/(mv^(2)//2)` Which is the distance of CLOSEST approach |
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