Question

In a Rutherford scattering experiment, when an alpha particle of mass \(m_1\) approaches a target nucleus of charge \(Ze\) and mass \(m_2\), the distance of the closest approach is \(d_0\). The energy of the particle is:

• A. directly proportional to \(Z\)
• B. inversely proportional to \(Z\)
• C. directly proportional to mass \(m_2\)
• D. directly proportional to \(m_1 m_2\)

Hint:

At the closest approach all kinetic energy becomes Coulomb potential energy, \(E = 2Ze^2/(4\pi\varepsilon_0 d_0)\). For fixed \(d_0\) this is proportional to \(Z\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
An alpha particle fired at a nucleus is repelled by the Coulomb force. At the distance of closest approach \(d_0\) the particle stops for an instant, so all its kinetic energy has become electrostatic potential energy.

Step 2: Key Formula or Approach:
Charge of the alpha particle is \(2e\). Charge of the nucleus is \(Ze\).
Energy of the particle \(E\) equals the potential energy at \(d_0\):
\[ E = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{d_0} = \frac{2Ze^2}{4\pi\varepsilon_0\, d_0} \]

Step 3: Read the dependence:
For the given distance \(d_0\), everything else in this expression is a constant except \(Z\). So \(E \propto Z\). A larger nuclear charge repels more strongly, so the particle needs more energy to reach the same \(d_0\).
The masses \(m_1\) and \(m_2\) do not appear in the expression at all.

Step 4: Check option (A):
\(E \propto Z\) is exactly what the formula shows. So (A) is TRUE.

Step 5: Check option (B):
Inverse proportion would mean a nucleus with more charge needs less energy, which is opposite to Coulomb repulsion. So (B) is FALSE.

Step 6: Check options (C) and (D):
The target is treated as heavy and at rest, and the closest approach depends only on the charges and the distance. The energy does not depend on \(m_2\) or on \(m_1 m_2\). So (C) and (D) are FALSE.

Final Answer:
The energy is directly proportional to \(Z\), so option (A) is correct. \[ \boxed{E = \frac{2Ze^2}{4\pi\varepsilon_0 d_0} \propto Z} \]