A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of `10fm` to the nucleus. The de - Broglie wavelength (in units of fm) of the proton at its start is take the proton mass, `m_p = 5//3xx10^(-27) kg, h//e = 4.2xx10^(-15) J-s//C`, `(1)/(4piepsilon_0) = 9xx10^9m //F, 1 fm = 10^(-15)`.

Explanation

Here, `q_(1)=e` and `q_(2)=120 e`
`r_(0)=10fm=10xx10^(-15)m=10^(-14)m`
`m_(p)=5/3xx10^(-27)kg, h/e=4.2xx10^(-15)Js//C`
`1/(4piepsilon_(0))=9xx10^(9)m//F`
Using the principal of coservation of energy
`(q_(1)q_(2))/(4piepsilon_(0)r_(0))=1/2mv^(2)=(p^(2))/(2m)=((h/lambda)^(2))/(2m)=(h^(2))/(2mlambda^(2))`
`lambda=sqrt((4piepsilon_(0)r_(0)h^(2))/(q_(1)q_(2) (2m)))`
`=sqrt(((10xx10^(-15))h^(2)xx3)/(9xx10^(9)e(120e)xx2xx5xx10^(-27)))`
`=sqrt((30xx10^(-15)(4.2xx10^(-15))^(2))/(9xx10^(9)e(120e)xx2xx5xx10^(-27)))`
`lambda=7xx10^(-15)m=7m`