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 \(\frac{1}{4\pi\epsilon_0}=9\times10^{9}\) SI units)

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

Hint:

The distance of closest approach gives an upper limit estimate for the size of the nucleus. Remember the conversion factor \(1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}\).

Answer 1

The Correct Option is C

The problem involves finding the distance of closest approach of an alpha particle to a gold nucleus. This is solved using energy conservation in electrostatics.

At the point of closest approach, the entire kinetic energy of the alpha particle is converted into electrostatic potential energy.

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Given:

• Energy of alpha particle: \(E = 7.7 \,\text{MeV}\)
• Atomic number of gold: \(Z = 79\)
• Charge of alpha particle: \(+2e\)
• \(\displaystyle \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \,\text{N·m}^2/\text{C}^2\)
• Charge of electron: \(e = 1.6 \times 10^{-19}\,\text{C}\)

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Formula:

At closest approach,

\[ E_{\text{kinetic}} = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r} \]

Solving for \(r\):

\[ r = \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^2}{E_{\text{kinetic}}} \]

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Calculation:

1. Convert energy from MeV to joules:

   \[ 7.7\,\text{MeV} = 7.7 \times 1.6 \times 10^{-13} = 1.232 \times 10^{-12}\,\text{J} \]

2. Substitute values:

   \[ r = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2} {1.232 \times 10^{-12}} \]

3. Evaluating:

   \[ r \approx 2.95 \times 10^{-14}\,\text{m} \]

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Final Answer:

\(\boxed{2.95 \times 10^{-14}\ \text{m}}\)