Modern Physics 4 Question 25
26. 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-Broglie wavelength (in units of $fm$ ) of the proton at its start is [Take the proton mass, $m _p=(5 / 3) \times 10^{-27} kg ; h / e=4.2 \times 10^{-15}$ $J-s / C$;
$\left.\frac{1}{4 \pi \varepsilon _0}=9 \times 10^{9} m / F ; 1 fm=10^{-15} m\right]$
(2013 Adv.)
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Solution:
$r=$ closest distance $=10 fm$.
From energy conservation, we have
$$ \begin{array}{rlrl} K _i+U _i & =K _f+U _f \\ \text { or } & & \\ \text { or } & & =0+\frac{1}{4 \pi \varepsilon _0} \cdot \frac{q _1 q _2}{r} \\ K & =\frac{1}{4 \pi \varepsilon _0} \cdot \frac{(120 e)(e)}{r} \end{array} $$
de-Broglie wavelength
$$ \lambda=\frac{h}{\sqrt{2 K m}} $$
Substituting the given values in above two equations, we get
$$ \lambda=7 \times 10^{-15} m=7 fm $$
