If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ___ m. (Atomic number of gold = 79 and $$\frac{1}{4\pi\epsilon _{0}}=9\times10^{9} \text{in SI units} )$$

An alpha particle with energy 7.7 MeV approaches a gold nucleus (Z = 79) and we need the closest distance of approach.

Apply energy conservation: at the closest distance all kinetic energy converts to electrostatic potential energy given by $$KE = \frac{1}{4\pi\epsilon_0} \cdot \frac{Z_1 Z_2 e^2}{d}$$ where $$Z_1 = 2$$, $$Z_2 = 79$$, and $$e = 1.6 \times 10^{-19}$$ C.

Since we solve for $$d$$ we write $$d = \frac{kZ_1 Z_2 e^2}{KE}$$ where $$k = 9 \times 10^9 \text{ N m}^2\text{/C}^2$$ and $$KE = 7.7 \times 10^6 \times 1.6 \times 10^{-19}$$ J.

Substituting these values yields $$d = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.7 \times 10^6 \times 1.6 \times 10^{-19}}.$$

The numerator is $$9 \times 10^9 \times 158 \times 2.56 \times 10^{-38} = 3639.17 \times 10^{-29} = 3.639 \times 10^{-26}$$ and the denominator is $$12.32 \times 10^{-13} = 1.232 \times 10^{-12}$$.

This gives $$d = \frac{3.639 \times 10^{-26}}{1.232 \times 10^{-12}} = 2.95 \times 10^{-14} \text{ m}.$$

Therefore the closest distance of approach is $$2.95 \times 10^{-14} \text{ m}.$$