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)
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)
(Atomic number of gold = 79 and \(\frac{1}{4\pi\epsilon_0}=9\times10^{9}\) SI units)
Show Hint
The distance of closest approach gives an upper limit estimate for the size of the nucleus. Remember the conversion factor \(1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}\).
Updated On: Mar 21, 2026
- \(2.95 \times 10^{-16}\)
- \(3.85 \times 10^{-14}\)
- \(2.95 \times 10^{-14}\)
- \(3.85 \times 10^{-16}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
An alpha particle approaching a nucleus is repelled by the Coulombic force. At the distance of closest approach (\(d\)), the entire initial kinetic energy (\(K\)) of the alpha particle is converted into electric potential energy (\(U\)). Step 2: Key Formula:
By conservation of energy: \[ K = U = \frac{1}{4\pi\epsilon_0} \frac{(Ze)(2e)}{d} \] Where:
\(K\) is the kinetic energy of the alpha particle.
\(Z\) is the atomic number of the target nucleus (Gold).
\(e\) is the elementary charge.
\(d\) is the distance of closest approach. Step 3: Detailed Calculation:
Given values:
\(K = 7.7 \, \text{MeV} = 7.7 \times 10^6 \times 1.6 \times 10^{-19} \, \text{J}\)
\(Z = 79\)
\(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\)
Charge of alpha particle \(q_{\alpha} = 2e\), Charge of nucleus \(q_{Au} = Ze\). Rearranging the formula for \(d\): \[ d = \frac{2Ze^2}{4\pi\epsilon_0 K} = \frac{2 Z e^2 (9 \times 10^9)}{K} \] Substitute the values: \[ d = \frac{2 \times 79 \times (1.6 \times 10^{-19})^2 \times (9 \times 10^9)}{7.7 \times 1.6 \times 10^{-13}} \] \[ d = \frac{2 \times 79 \times 1.6 \times 10^{-38} \times 9 \times 10^9}{7.7 \times 10^{-13}} \] \[ d = \frac{2 \times 79 \times 1.6 \times 9}{7.7} \times 10^{-38+9+13} \] \[ d = \frac{2275.2}{7.7} \times 10^{-16} \] \[ d \approx 295.48 \times 10^{-16} \, \text{m} \] \[ d \approx 2.95 \times 10^{-14} \, \text{m} \] Step 4: Final Answer:
The closest distance is \(2.95 \times 10^{-14}\) m.
An alpha particle approaching a nucleus is repelled by the Coulombic force. At the distance of closest approach (\(d\)), the entire initial kinetic energy (\(K\)) of the alpha particle is converted into electric potential energy (\(U\)). Step 2: Key Formula:
By conservation of energy: \[ K = U = \frac{1}{4\pi\epsilon_0} \frac{(Ze)(2e)}{d} \] Where:
\(K\) is the kinetic energy of the alpha particle.
\(Z\) is the atomic number of the target nucleus (Gold).
\(e\) is the elementary charge.
\(d\) is the distance of closest approach. Step 3: Detailed Calculation:
Given values:
\(K = 7.7 \, \text{MeV} = 7.7 \times 10^6 \times 1.6 \times 10^{-19} \, \text{J}\)
\(Z = 79\)
\(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\)
Charge of alpha particle \(q_{\alpha} = 2e\), Charge of nucleus \(q_{Au} = Ze\). Rearranging the formula for \(d\): \[ d = \frac{2Ze^2}{4\pi\epsilon_0 K} = \frac{2 Z e^2 (9 \times 10^9)}{K} \] Substitute the values: \[ d = \frac{2 \times 79 \times (1.6 \times 10^{-19})^2 \times (9 \times 10^9)}{7.7 \times 1.6 \times 10^{-13}} \] \[ d = \frac{2 \times 79 \times 1.6 \times 10^{-38} \times 9 \times 10^9}{7.7 \times 10^{-13}} \] \[ d = \frac{2 \times 79 \times 1.6 \times 9}{7.7} \times 10^{-38+9+13} \] \[ d = \frac{2275.2}{7.7} \times 10^{-16} \] \[ d \approx 295.48 \times 10^{-16} \, \text{m} \] \[ d \approx 2.95 \times 10^{-14} \, \text{m} \] Step 4: Final Answer:
The closest distance is \(2.95 \times 10^{-14}\) m.
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