Question

In a geiger - marsden experiment. Find the distance of closest approach to the nucleus of a 7.7 me v `alpha`- particle before it comes momentarily to rest and reverses its direction. (z for gold nucleus = 79) .
A. 10 fm
B. 20 fm
C. 40 fm
D. 10 fm

Answer 1

Correct Answer - C
Let d be the distance of closest approach then by the conservation of energy. Initial kinetic energy of incoming `alpha`-particle K.
= Final electric potential energy U of the sysyem
As `K=(1)/(4piepsilon_0)xx((2e)(Ze))/(d)therefored=(1)/(4piepsilon_0)(2Ze^2)/(K)`....(i)
Here,
`(1)/(4piepsilon_0)=9xx10^9Nm^2C^(-2)`,`Z=79,e=1.6xx10^(-19)C`.
`K =7.7MeV=7.7xx10^6xx1.6xx10^(-19)J=1.2xx10^(-12)J` Substituting these values in (i)
`d= (2xx9xx10^(9)xx(1.6xx10^(-19))^2xx79)/(1.2xx10^(-12))`
`d=3xx10^(-14)m=30fm` (`because 1 fm=10^(-15)m`)