A proton is fired from very far away towards a nucleus with charge Q=120e, where e is the electronic charge. It makes a closest approach of 10 fm 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)/(3))xx10^(-27)kg,(h)/(e)=4.2xx10^(-15)Js//C,(1)/(4piepsilon_(0))=9xx10^(9)m//F,1fm=10^(-15)m)`

Answer

`7`

Explanation

`((9xx10^(9))(120e)(e))/(10xx10^(-15))=(p^(2))/(2m)`
`lambda=(h)/(p) :. p^(2)=(h^(2))/(lambda^(2))`
`2((5)/(3)xx10^(-27))10^(15)(9xx10^(9))(12)e^(2)=(h^(2))/(2mlambda^(2))`
`(120)(3)10^(-27+15+9) " " lambda^(2)=(4.2)^(2)xx10^(-30)`
`lambda^(2)=(4.2xx4.2xx10^(-30))/(360xx10^(-3))=(42xx42)/(360)xx10^(-29)`
`7^(2)xx10^(-30) " " lambda=7xx10^(-15)m`
`=7 fm`