An particle of energy 5 MeV is scattered through 180° by gold nucleus. The distance of closest approach is of the order of

Options

A. `10^(-12) cm `
B. `10^(-16) cm `
C. `10^(-10)` cm
D. `10^(-14)` cm

Answer

A

Explanation

At the distance of closest approach d , Kinetic energy of `alpha ` - particle
= Potential energy of `alpha - ` particle and gold nucleus
` i.e ., K = 1/(4piepsilon) ((2e)(Ze))/d = (2Ze^(2))/(4piepsilon_(0)d)`
or ` d = (2Ze^(2))/(4piepsilon_(0)K)`
Here , `K = 5 Mev = 5xx 1.6 xx10^(-13) J `
`( :' 1 MeV = 1.6 xx10^(13) J ) `
For gold , Z = 79
` :. " " d = ((2)(9xx10^(9)" Nm"^(-2)C^(-2))(79) (1.6 xx10^(-19)C)^(2))/((5xx1.6 xx10^(-13)J))`
` = 4.55 xx10^(-14) m approx 10^(-12) cm `