Question

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

• A. 1.72 nm
• B. 6.2 nm
• C. 16.8 nm
• D. 0.2 nm

Hint:

The closest approach in electrostatics depends on the kinetic energy of the incoming particle and the electrostatic potential energy at the point of closest approach.

Answer 1

The Correct Option is A

To find the distance of closest approach of an \(\alpha\)-particle to a fixed gold nucleus, we can use the principle of conservation of energy. The kinetic energy of the \(\alpha\)-particle is converted into electrostatic potential energy at the distance of closest approach.

The initial kinetic energy \((K.E)\)of the \(\alpha\)-particle is 7.7 MeV. The electrostatic potential energy \((U)\)at the closest distance \((r)\)is given by:

\(U = \frac{k \cdot Z_1 \cdot Z_2 \cdot e^2}{r}\)

Where:

• \(k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\)(Coulomb's constant)
• \(Z_1 = 2\)(atomic number of \(\alpha\)-particle)
• \(Z_2 = 79\)(atomic number of gold nucleus)
• \(e = 1.6 \times 10^{-19} \, \text{C}\)(elementary charge)
• \(r\)is the distance of closest approach

By conservation of energy, we have:

\(K.E = U\)

\(7.7 \, \text{MeV} = \frac{k \cdot 2 \cdot 79 \cdot e^2}{r}\)

First, convert the kinetic energy from MeV to joules:

\(1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}\)

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

Substituting the values into the equation:

\(1.232 \times 10^{-12} = \frac{8.99 \times 10^9 \cdot 2 \cdot 79 \cdot (1.6 \times 10^{-19})^2}{r}\)

This simplifies to:

\(r = \frac{8.99 \times 10^9 \cdot 2 \cdot 79 \cdot (1.6 \times 10^{-19})^2}{1.232 \times 10^{-12}}\)

Calculating:

\(r = \frac{8.99 \times 10^9 \times 2 \times 79 \times (2.56 \times 10^{-38})}{1.232 \times 10^{-12}}\)

\(r \approx 1.72 \times 10^{-9} \, \text{m}\)

Thus, the distance of closest approach is approximately \(1.72 \, \text{nm}\).

Therefore, the correct answer is 1.72 nm.