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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 $10 \ fm$ to the nucleus. The de Brogle wavelength (in units of $fm$ ) of the proton at its start is :
(take the proton mass, $m _0=(5 / 3) \times 10^{-27} kg , h / e =4.2 \times 10^{-15} J / s / C ; \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 m / F ; 1 fm =10^{-15} m$ )
$7$
$8$
$9$
$1$
Solution
$Image$
$\frac{\left(9 \times 10^9\right)(120 e )( e )}{10 \times 10^{-15}}=\frac{ p ^2}{2 m } $
$\lambda=\frac{ h }{ p } \quad \therefore p ^2=\frac{ h ^2}{\lambda^2} $
$2\left(\frac{5}{3} \times 10^{-27}\right) 10^{15}\left(9 \times 10^9\right)(12)^2=\frac{ h ^2}{2 m \lambda^2} $
$(120)(3) 10^{-27+15+9} $$ \lambda^2=(4.2)^2 \times 10^{-30} $
$\lambda^2=\frac{4.2 \times 4.2 \times 10^{-30}}{360 \times 10^{-3}} $$ =\frac{42 \times 42}{360} \times 10^{-29} $
$=7^2 \times 10^{-30} $$ \lambda=7 \times 10^{-15} m $
$=7 fm $
