Question

A \(7.9\ \text{MeV}\) \(\alpha\)-particle scatters from a target material of atomic number \(79\). From the given data, the estimated diameter of the nuclei of the target material is (approximately) _________ m. \[ \left[\frac{1}{4\pi\varepsilon_0}=9\times10^9\ \text{Nm}^2\text{/C}^2 \text{ and electron charge }=1.6\times10^{-19}\ \text{C}\right] \]

• A. \(1.69\times10^{-12}\)
• B. \(1.44\times10^{-13}\)
• C. \(2.88\times10^{-14}\)
• D. \(5.76\times10^{-14}\)

Hint:

The diameter of a nucleus can be estimated using the distance of closest approach formula from Rutherford scattering.

Answer 1

The Correct Option is C

Step 1: Use the distance of closest approach formula.
For Rutherford scattering, \[ d = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{E} \] where \(Z=79\), \(e=1.6\times10^{-19}\ \text{C}\), and \(E=7.9\ \text{MeV}=7.9\times10^6\times1.6\times10^{-19}\ \text{J}\).
Step 2: Substitute the values.
\[ d = 9\times10^9 \times \frac{2\times79\times(1.6\times10^{-19})^2} {7.9\times10^6\times1.6\times10^{-19}} \] \[ d \approx 2.88\times10^{-14}\ \text{m} \] Final Answer: \[ \boxed{2.88\times10^{-14}\ \text{m}} \]