Question

An alpha particle of energy \( K \, \text{MeV} \) is moving towards a nucleus of atomic number \( Z \). The distance of closest approach of the alpha particle to the nucleus in meters is

• A. \( 7.2 \times 10^{-16} \, \frac{Z}{K} \)
• B. \( 3.84 \times 10^{-16} \, \frac{Z}{K} \)
• C. \( 14.4 \times 10^{-16} \, \frac{Z}{K} \)
• D. \( 28.8 \times 10^{-16} \, \frac{Z}{K} \)

Hint:

The distance of closest approach of an alpha particle to a nucleus depends on the energy of the particle and the atomic number of the nucleus.

Answer 1

The Correct Option is D

The distance of closest approach \( r \) of an alpha particle moving towards a nucleus is given by: \[ r = \frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{K} \] where \( Z \) is the atomic number of the nucleus, \( e \) is the charge of the electron, and \( K \) is the energy of the alpha particle. By substituting the known values and rearranging the equation, we find that the distance of closest approach is \( 28.8 \times 10^{-16} \, \frac{Z}{K} \). Thus, the correct answer is \( \boxed{28.8 \times 10^{-16} \, \frac{Z}{K}} \).