Question

An $\alpha$ particle approaches the target nucleus of copper (Z = 29) in such a way that the value of impact parameter is zero. The distance of closest approach will be

Answer 1

Answer:

     (KE​)α= 1/2 mv^2

             = 1/4π​​ε0 * ZeZe / d

             =Ze^2/4π​​ε0d * d

             =2Ze^2 / 4π​​ε0k

given Z = 29 (Cu)

    (KE)​α=2*29*e^2 / 4π​​ε0k

            =29e^2 / 2π​​ε0d

        d  = 2π​​ε0 (KE​)α / 29e^2

2π​​ε0(KE)​α/29e^2