Atomic StructureHard

Question

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 $\frac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9}$ in SI units)

Options

A.$2.95 \times 10^{- 14}$
B.$2.95 \times 10^{- 16}$
C.$3.85 \times 10^{- 16}$
D.$3.85 \times 10^{- 14}$

Solution

Energy conservation

$$\begin{matrix} & K_{i} + U_{i} = K_{f} + U_{f} \\ & 7.7 \times 10^{6} \times 1.6 \times 10^{- 19} + 0 \\ & \ = 0 + \frac{9 \times 10^{9}\left( 1.6 \times 10^{- 19} \right)\left( 79 \times 1.6 \times 10^{- 19} \right)}{r} \\ & r = 2.95 \times 10^{- 14} \end{matrix}$$

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