Alpha Particle Of Energy 5 Mev Is Scattered Through 180в° By Fixed Uranium Nucleus

An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest … cm (C) 10-12 cm…

Answers:

Correct option (C) 10-12 cm

 Explanation:

 From conservation of mechanical energy 

 Decrease in kinetic energy = increase in potential energy

Or 1/4π ε((Ze) (ze))/rmin = 5 MeV = 5 ×1.6 × 10-13 J

∴ r min = 1/4 π ε0  2Ze2/5 × 1.6 × 10-13

= ((9 × 109) (2) (92) (1.6 × 10-19)2)/(5 × 1.6 × 10-13)

= 5.3 × 10-14 m

= 5.3 × 10-12 cm

i.e., rmin is of the order of 10-12 cm