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:

For head-on Rutherford scattering problems: \[ r_{\text{min}} = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{K} \] Higher kinetic energy means the particle can approach closer to the nucleus, while a larger nuclear charge increases repulsion.

Answer 1

The Correct Option is A

Concept: When an $\alpha$-particle approaches a heavy nucleus, it experiences electrostatic repulsion. As it comes closer, its kinetic energy decreases. At the distance of closest approach, the $\alpha$-particle momentarily stops and its entire kinetic energy is converted into electrostatic potential energy. The electrostatic potential energy between two point charges is: \[ U = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{r} \]
Step 1: Identify given quantities Charge of $\alpha$-particle: \[ Z_1 = 2 \] Atomic number of gold nucleus: \[ Z_2 = 79 \] Kinetic energy of $\alpha$-particle: \[ K = 7.7 \text{ MeV} = 7.7 \times 1.6 \times 10^{-13} \text{ J} \]
Step 2: Apply conservation of energy At closest approach, \[ K = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{r} \] Rearranging, \[ r = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{K} \]
Step 3: Substitute values \[ r = \frac{(9 \times 10^9)(2)(79)(1.6 \times 10^{-19})^2}{7.7 \times 1.6 \times 10^{-13}} \]
Step 4: Calculate \[ r \approx 1.72 \times 10^{-9} \text{ m} \]
Step 5: Final answer \[ r = 1.72 \text{ nm} \]