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

To find the distance of closest approach when an \( \alpha \)-particle is projected towards a stationary target nucleus, we use the principle of conservation of energy. In this case, the initial kinetic energy (\( KE \)) of the \( \alpha \)-particle is entirely converted into electrostatic potential energy at the point of closest approach.

The formula for the distance of closest approach \( d \) is given by:

d = \frac{k \cdot Z \cdot e^2}{KE}

where:

• \( k \) is Coulomb's constant, \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)
• \( Z \) is the atomic number of the target nucleus
• \( e \) is the charge of an electron, \( 1.6 \times 10^{-19} \, \text{C} \)
• \( KE \) is the kinetic energy of the \( \alpha \)-particle, given in electronvolts (eV)

Substitute the given values into the equation:

• \( Z = 79 \)
• \( KE = 7.9 \, \text{MeV} = 7.9 \times 10^6 \, \text{eV} = 7.9 \times 10^6 \times 1.6 \times 10^{-19} \, \text{J} \)

d = \frac{(8.99 \times 10^9) \times 79 \times (1.6 \times 10^{-19})^2}{7.9 \times 10^6 \times 1.6 \times 10^{-19}}

Simplifying the expression:

d = \frac{(8.99 \times 10^9) \times 79 \times (1.6 \times 10^{-19})}{7.9 \times 10^6}

On further calculation:

d = \frac{1.803584 \times 10^{-27}}{1.264 \times 10^{-12}}

d = 2.88 \times 10^{-14} \, \text{m}

Thus, the distance of closest approach is \( 2.88 \times 10^{-14} \, \text{m}.

Therefore, the correct answer is: \( 2.88 \times 10^{-14} \)