A) What initial kinetic energy must an alpha particle have ifit is to approach a

Question

A) What initial kinetic energy must an alpha particle have ifit is to approach a stationary gold nucleus to within a distance of24.2 fm?
B) If the initial speed of the alpha particle isreduced by a factor of 2.70, by what factor is the distance ofclosest approach changed? Enter positive number for increase ofdistance or negative number for decrease.
A) What initial kinetic energy must an alpha particle have ifit is to approach a stationary gold nucleus to within a distance of24.2 fm?
B) If the initial speed of the alpha particle isreduced by a factor of 2.70, by what factor is the distance ofclosest approach changed? Enter positive number for increase ofdistance or negative number for decrease.

Explanation / Answer

A)the kinetic energy the alpha particle must have is K = (h * c/) --------(1) h = 6.63 * 10-34 J-s,c = 3 * 108 m/s and =24.2 fm = 24.2 * 10-15 m the speed of the alpha particle be u therefore we get (1/2)m * u2 = (h * c/) or u = (2h * c/m * )1/2 ----------(2) m = 1.67 * 10-27 kg when the speed of the alpha particle is reduced by a factor of(u/2.70) then let the distance of closest approach be1 therefore we get (1/2)m * (u/2.70)2 = (h *c/1) or 1 = (14.58 * hc/m * u2) m = 1.67 * 10-27 kg and the value of u is obtainedfrom equation (2).

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