Question

Determine the distance of the closest approach when an alpha particle of kinetic energy 3.95 MeV approaches a nucleus of Z = 79, stops and reverses its directions.

Answer 1

Let 'r' be centre to centre distance between α-particle when α particle is at the stopping point then K = `1/(4piepsilon_0) ((Ze)(2e))/r`

Given: K = 3.95 MeV

= `3.95 xx 10^6 xx 1.6 xx 10^-19` J

⇒ `r = 1/(4piepsilon_0) (2Ze^2)/K`

= `(9 xx 10^9 xx 2 xx 79 xx (1.6 xx 10^-19)^2)/(3.95 xx 10^6 xx 1.6 xx 10^-19)`

= `(9 xx 2 xx 79 xx 1.6 xx 10^-19 xx 10^9)/(3.95 xx 10^6)`

= `576 xx 10^-16` m

`r = 5.76 xx 10^-14` m