41.
Alpha particles having kinetic energy of 1.8 MeV each are incident on a thin gold foil, form a large distance. Applying the principle of conservation of energy, find the closest distance of approach of the alpha particle from the gold nucleus.
Kinetic energy of α-particle = Potential energy of repulsion between α-particle and nucleus.

= 1264 x 10 -16
r0 = 1.264 x 10-13m
The distance of closest approach is 1.264 x 10-13 m.
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