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Physics Question on Atoms

Alpha particles of kinetic energy 7.7 MeV7.7 \,MeV are being scattered by the nucleus of gold which has 7979 electrons. The distance of closest approach of the alpha particles is (Take 14 π ε0=9×109 MKS) \frac{1}{4\,\pi\, \varepsilon_{0} = 9 \times 10^{9} \, MKS} )

A

4×10−14m4\times 10^{-14} m

B

30×10−15m30 \times 10^{-15} m

C

10×10−15m10\times 10^{-15} m

D

7.9×10−14m7.9\times 10^{-14} m

Answer

30×10−15m30 \times 10^{-15} m

Explanation

Solution

We know that
14πε0⋅(Ze)(2e)r0=K\frac{1}{4\pi\varepsilon_{0}}\cdot\frac{\left(Ze\right)\left(2e\right)}{r_{0}} = K
9×109×79×(1.6×10−19)2r0×29\times 10^{9} \times\frac{ 79\times \left(1.6 \times 10^{-19}\right)^{2}}{r_{0}} \times 2
=7.7×106×1.6×10−19= 7.7 \times 10^{6} \times 1.6 \times 10^{-19}
r0=9×109×79×(1.6×10−19)2×27.7×106×1.6×10−19r_{0} =\frac{9\times 10^{9} \times 79 \times\left(1.6\times 10^{-19}\right)^{2} \times 2}{7.7\times 10^{6} \times 1.6\times 10^{-19}}
=2.9×10−14= 2.9\times 10^{-14}
r0=3.0×10−14r_{0} = 3.0\times 10^{-14}
=30×10−15m= 30\times 10^{-15} m