An particle approaches the target nucleus of copper (Z = 29) in such

1.	An 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» ANSWER: (KE)α= 1/2 mv^2 = 1/4πε0 * ZeZe / d =Ze^2/4πε0d * d =2Ze^2 / 4πε0k given Z = 29 (CU) (KE)α=229e^2 / 4πε0k =29e^2 / 2πε0d d = 2πε0 (KE)α / 29e^2 2πε0(KE)α/29e^2

An 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»

(KE)α= 1/2 mv^2

= 1/4πε0 * ZeZe / d

=Ze^2/4πε0d * d

=2Ze^2 / 4πε0k

given Z = 29 (CU)

(KE)α=2*29*e^2 / 4πε0k

=29e^2 / 2πε0d

2πε0(KE)α/29e^2