Question

An \( \alpha \)-particle with KE 7.9 MeV is projected towards a stationary target nucleus of \( Z = 79 \). Find the distance of closest approach.

• A. \( 1.44 \times 10^{-14} \)
• B. \( 2.88 \times 10^{-14} \)
• C. \( 1.44 \times 10^{-15} \)
• D. \( 2.88 \times 10^{-15} \)

Hint:

The formula for the closest approach is useful for determining the interaction distance between charged particles and atomic nuclei.

Answer 1

The Correct Option is B

Step 1: Formula for closest approach.
The distance of closest approach \( r \) for an \( \alpha \)-particle is given by the formula: \[ r = \frac{1.44 \times 10^{-14} \, \text{m} \times Z}{KE \, (\text{MeV})} \] where \( Z \) is the atomic number of the target nucleus and \( KE \) is the kinetic energy of the \( \alpha \)-particle in MeV. Step 2: Apply the given values.
For \( Z = 79 \) and \( KE = 7.9 \, \text{MeV} \), we calculate the distance of closest approach: \[ r = \frac{1.44 \times 10^{-14} \times 79}{7.9} \, \text{m} = 2.88 \times 10^{-14} \, \text{m} \] Step 3: Conclusion.
The distance of closest approach is \( 2.88 \times 10^{-14} \, \text{m} \). Final Answer: \[ \boxed{2.88 \times 10^{-14}} \]