Question

An α-particle of 10 MeV is moving forward for a head on collision. What will be the distance of closest approach from the nucleus of atomic number Z = 50?choose the correct option from the given options.
(A) 1.44*10⁻¹⁴m
(B) 2.88*10⁻¹⁴m
(C) 0.53*10⁻¹⁰m
(D) 0.53*10⁻¹⁰/50m

Answer 1

With the given values you have to derive right and best solution for the customer and it assure to provide better choice for the client and wish to move with no risk and trouble of it .

I think you have to equate kinetic and potential energy ie 1/2mv2=1/4pi€×2Ze2/r r is the distance of closest approach by substituting the values you get r=14.4×10-15 m

Answer 2

Holla ^_^


☸ Required Answer is - B) 2.88*10⁻¹⁴m


Vielen Dank ♥