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 f m to the nucleus. The de Broglie wavelength (in units of f m) of the proton at its start is (take the proton mass,
   
$m_p=\frac{5}{3}\times {10}^{-27}kg;\frac{h}{e}=4.2\times {10}^{-15}J.s/C;\frac{1}{4\pi {\in }_0}=9\times {10}^9$

$m/F;1fm={10}^{-15}m)$