$$7.9 MeV \alpha - \text{particle}$$ scatters from a target material of atomic muuber 79. From the given data the estimated diameter of nuclei of the target material is (approximately) ___m.

$$\left[ \frac{1}{4\pi \epsilon_{o}}=9\times 10^{9} Nm^{2}/c^{2} \text{ and electron change}=1.6\times 10^{-19}C \right ]$$

Estimate the diameter of a nucleus (Z = 79) from alpha-particle scattering data.

When an alpha particle (charge $$2e$$, kinetic energy $$KE$$) approaches a nucleus head-on, all its kinetic energy converts to electrostatic potential energy at the distance of closest approach ($$r_0$$):

$$ KE = \frac{1}{4\pi\epsilon_0} \cdot \frac{(Ze)(2e)}{r_0} $$

The diameter $$d_0 = 2r_0$$. Solving for $$r_0$$ gives

$$ r_0 = \frac{1}{4\pi\epsilon_0} \cdot \frac{2Ze^2}{KE} $$

With $$KE = 7.9$$ MeV $$= 7.9 \times 10^6 \times 1.6 \times 10^{-19}$$ J $$= 12.64 \times 10^{-13}$$ J $$= 1.264 \times 10^{-12}$$ J, we substitute numerical values into the expression for $$r_0$$:

$$ r_0 = 9 \times 10^9 \times \frac{2 \times 79 \times (1.6 \times 10^{-19})^2}{1.264 \times 10^{-12}} $$

The numerator is $$2 \times 79 = 158$$ and $$(1.6 \times 10^{-19})^2 = 2.56 \times 10^{-38}$$, so

$$158 \times 2.56 \times 10^{-38} = 404.48 \times 10^{-38} = 4.0448 \times 10^{-36}$$ and

$$9 \times 10^9 \times 4.0448 \times 10^{-36} = 36.4 \times 10^{-27} = 3.64 \times 10^{-26}$$ giving

$$r_0 = \frac{3.64 \times 10^{-26}}{1.264 \times 10^{-12}} = 2.88 \times 10^{-14}$$ m.

Therefore the diameter is $$d_0 = 2r_0 = 2 \times 2.88 \times 10^{-14} = 5.76 \times 10^{-14}$$ m, which corresponds to Option C: $$5.76 \times 10^{-14}$$ m.