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)`.

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`