Question

An \( \alpha \)-particle having kinetic energy \(7.7\,\text{MeV}\) is approaching a fixed gold nucleus (atomic number \(Z = 79\)). Find the distance of closest approach.

• A. \(1.72\,\text{nm}\)
• B. \(6.2\,\text{nm}\)
• C. \(16.8\,\text{nm}\)
• D. \(0.2\,\text{nm}\)

Hint:

For closest approach problems: \begin{itemize} \item Assume head-on collision \item Equate kinetic energy to electrostatic potential energy \item Use \(1.44\,\text{MeV·fm}\) for quick nuclear calculations \end{itemize}

Answer 1

The Correct Option is D

The problem involves calculating the distance of closest approach of an \( \alpha \)-particle, which is a fundamental problem in electrostatics and nuclear physics. The kinetic energy of the \( \alpha \)-particle is converted entirely into electric potential energy at the distance of closest approach.

Given:

• Kinetic Energy (\( K.E. \)) of the \( \alpha \)-particle is \( 7.7 \, \text{MeV} \). 
• Atomic number (\( Z \)) of gold is \( 79 \).
• Charge of an \( \alpha \)-particle (\( q \)) is \( 2e \). \( e \) is the elementary charge (\( 1.6 \times 10^{-19} \, \text{C} \)).
• The potential energy at the distance of closest approach is given by the formula:

\(U = \dfrac{1}{4\pi\epsilon_0} \cdot \dfrac{qQ}{r}\)

Where:

• \(Q = Ze\) (charge of the gold nucleus)
• \(r\) is the distance of closest approach
• {K.E.} = U

 

Thus,

\(7.7 \times 10^6 \times 1.6 \times 10^{-19} = \dfrac{1}{4\pi\epsilon_0} \dfrac{2 \cdot 79 \cdot (1.6 \times 10^{-19})^2}{r}\)

Substituting for \(\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2\), we get:

\(7.7 \times 1.6 \times 10^{-13} = 9 \times 10^9 \times \dfrac{2 \times 79 \times (1.6 \times 10^{-19})^2}{r}\)

\(r = \dfrac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{7.7 \times 1.6 \times 10^{-13}}\)

Calculating \(r\):

\(r \approx \dfrac{9 \times 10^9 \times 2 \times 79 \times 2.56 \times 10^{-38}}{1.232 \times 10^{-12}}\)

\(r \approx 2.06 \times 10^{-10} \text{m} = 0.206 \, \text{nm}\)

Rounding to one decimal place, the distance of closest approach is approximately \(0.2 \, \text{nm}\).

Therefore, the correct answer is \( 0.2 \, \text{nm} \).