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

`hv=W+13.6eV` and in second case `KE_(max)=(12420)/(1215)=10.2eV`
` therefore5/6hv=W+10.2eVRightarrowv=5xx10^(15)H_(z)`