Explanation
Given here,
$$q_{1} = 120 \times e$$
$$q_{2} = e $$ ( charge on proton)
It is based on conservation of energy of proton.
$$\dfrac{1}{2}mv^2 = k \dfrac{q_1q_2}{R}$$
$$v = \sqrt{\dfrac{2Kq_1q_2}{mR}}$$
Also, we know, de’broglie wavelength,
$$\lambda =\dfrac{h}{mv}$$
$$\lambda = \dfrac{h\sqrt{R}}{\sqrt{2mKq_1q_2}}\\$$
Thus,
$$\lambda = \dfrac{h\sqrt{10\times10^{-15}}}{\sqrt{2\times\dfrac{5}{3}\times10^{-27}\times9\times10^9\times120\times e^2}}\\$$
$$\lambda =\dfrac{\dfrac{h}{e}\times10^{-7}}{6\times10^{-8}}\\$$
$$\lambda =\dfrac{4.2\times10^{-15}\times10^{-7}}{6\times10^{-8}}\\$$
$$\lambda =7\times10^{-15}\ m = 7\ fm \\$$
Option A is correct.
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