Question

An \( \alpha \)-particle of 10 MeV collides head-on with a copper nucleus (\( Z = 29 \)) and is deflected back. The minimum distance of approach between the centers of the two is:

• A. \( 8.4 \times 10^{-15} \) cm
• B. \( 8.4 \times 10^{-15} \) m
• C. \( 4.2 \times 10^{-15} \) m
• D. \( 4.2 \times 10^{-15} \) cm

Hint:

Minimum distance of approach is calculated using electrostatic potential energy conversion.

Answer 1

The Correct Option is B

Step 1: Formula for minimum distance of approach.
The minimum distance of approach \( r_{{min}} \) between an \( \alpha \)-particle and a nucleus can be found using the formula:
\[ r_{{min}} = \frac{K Z_1 Z_2 e^2}{2E} \] where:
- \( K \) is Coulomb's constant, \( K = 9 \times 10^9 \, \mathrm{N m^2/C^2} \)
- \( Z_1 \) and \( Z_2 \) are the atomic numbers of the two particles (\( Z_1 = 2 \) for \( \alpha \)-particle, \( Z_2 = 29 \) for copper)
- \( e \) is the elementary charge, \( e = 1.6 \times 10^{-19} \, \mathrm{C} \)
- \( E \) is the kinetic energy of the \( \alpha \)-particle, \( E = 10 \, {MeV} = 10 \times 10^6 \times 1.6 \times 10^{-13} \, {J} \)
Step 2: Substituting values.
Substitute the known values into the formula: \[ r_{{min}} = \frac{(9 \times 10^9) \times (2) \times (29) \times (1.6 \times 10^{-19})^2}{2 \times (10 \times 10^6 \times 1.6 \times 10^{-13})} \] Step 3: Simplification.
Simplifying the expression:
\[ r_{{min}} = \frac{(9 \times 10^9) \times (58) \times (2.56 \times 10^{-38})}{(3.2 \times 10^{-6})} \] \[ r_{{min}} = \frac{(9 \times 58 \times 2.56) \times 10^{-29}}{3.2 \times 10^{-6}} = \frac{1345.92 \times 10^{-29}}{3.2 \times 10^{-6}} = 8.4 \times 10^{-15} \, {m} \] Thus, the minimum distance of approach is \( 8.4 \times 10^{-15} \, {m} \).