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 $10 f m$ to the nucleus. The de Broglie wavelength (in units $f m$ ) of the proton at its start is: (take the proton mass, $m_{p} = (\frac{5}{3}) \times 10^{- 27} k g; h / e = 4.2 \times 10^{- 15} J . s / C; \frac{1}{4 π ε_{0}} = 9 \times 10^{9} m / F; 1 f m = 10^{- 15} m$

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7 $f m$

5 $f m$

2 $f m$

3 $f m$

answer is A.

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Detailed Solution

Los of K.E of proton = Gain of P.E of the proton - nucleus system

$\frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r} ∴ \frac{p^{2}}{2 m} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r} ∴ \frac{1}{2 m} (\frac{h^{2}}{λ^{2}}) = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r} ∴ λ = \sqrt{\frac{4 π ε_{0} r . h^{2}}{q_{1} q_{2} (2 m)}} = 7 f m$