Question

Alpha particles which have a charge +2e and mass 6.64 x 10-27 kg are initially at rest and are fired directly at a stationary platinum nucleus (charge +78e) at a speed of 2.75 x 10 m/s. Determine the distance of closest approach (in fm) before the alpha particle reverses direction. Assume the platinum nucleus remains stationary. Assume the alpha particles are initially very far from a stationary platinum nucleus. fm

Answer 1

Charge of alpha particles
91 = +2e
, mass of alpha particles
m1 = 6.64424 * 10-27 kg

Charge of platinum

92 = +78e

Initial velocity of alpha particles

V1 = 2.75 * 107 m/s

Initial velocity of platinum is zero.

At the closest distance of approach , the alpha particle stops momentarily, so its final velocity is zero at the moment.

Conserving the energy of alpha particles and platinum before and after collision,

PE; + KEY = PE; + KEF

$0+\frac{1}{2}m_1v_1^2=\left ( \frac{78e*2e}{4\pi\epsilon_0 r}\right )+0$

Τ = 2 * 78e * 2e 4περm,υ?

312(1.6 + 10-19) 4F(8.85 + 10-12)(6.64424 + 10-27)(2.75 + 10752 r


T = 1.429 * 10-14 m

( )

1 fm = 10-15 m

r = 14.29 fm

Distance of closest approach is

r = 14.29 fm