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

To solve this problem, we need to estimate the diameter of the nuclei using the scattering data provided. The key principle involved is the Rutherford scattering formula and potential energy consideration when the closest approach is achieved.

Let's use the formula for the distance of closest approach \( b \) in an alpha particle scattering in a Coulombic field:

\[ b = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2Ze^2}{K} \]

where:

• \(Z\) is the atomic number of the target nucleus, which is 79 for gold.
• \(e = 1.6 \times 10^{-19} \, \text{C}\) is the charge of the electron.
• \(K = 7.9\, \text{MeV} = 7.9 \times 1.6 \times 10^{-13}\, \text{J}\) is the kinetic energy.
• \(\frac{1}{4\pi \varepsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\).

Substituting these values into the formula:

\[ b = \left(9 \times 10^9 \, \text{N m}^2/\text{C}^2 \right) \cdot \left(\frac{2 \cdot 79 \cdot \left(1.6 \times 10^{-19}\right)^2}{7.9 \cdot 1.6 \times 10^{-13}}\right) \]

Calculating further:

• Atomic number for the target, \(Z = 79\)

\[ b = 9 \times 10^9 \times \frac{2 \times 79 \times 2.56 \times 10^{-38}}{1.264 \times 10^{-12}} \]

\[ b = 9 \times 10^9 \times \frac{404.48 \times 10^{-38}}{1.264 \times 10^{-12}} \]

\[ b \approx 9 \times 10^9 \times 3.203 \times 10^{-14} \approx 2.88 \times 10^{-14} \, \text{m} \]

Thus, the estimated diameter of the nuclei of the target material is approximately \(2.88 \times 10^{-14} \, \text{m}\).

Conclusion: The correct answer is \(2.88 \times 10^{-14} \, \text{m}\), which matches the provided option.