Plz solve problem 1 1 An alpha particle of energy 5 MeV is scattered through 180 by a - Physics - Atoms

Question

Plz solve problem 1
1 An alpha particle of energy 5 MeV is scattered through 180°
by a fixed uranium nucleus The distance of closest approach is of
the order of
(a) 1 A ˙ (b) 10-10 cm
(c) 10- 12 cm (d) 10-15 cm

Chandra Prakash · Accepted Answer

Dear student
at the closest distance all of the KE of alpha particle gets
converted into potentialenergy

1mv22=5MeV=kQq/rhere Q= charge on uranium=92e q= charge on alpha paricle = 2e 5*106eV=9*109*92*2*e*e/r r= 9*109*92*2*e5*106=9*109*92*2*1
6*10-195*106=2 94*10-14m2 94*10-12cm

Plz solve problem 1 1. An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of (a) 1 A ˙ (b) 10-10 cm (c) 10- 12 cm (d) 10-15 cm

Plz solve problem 1
1. An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of
(a) 1 $\dot{A}$ (b) 10^-10 cm
(c) 10^- ¹² cm (d) 10^-15 cm