Alpha particle scattering nuclear diameter estimation
Q. 7.9 MeV $\alpha$-particle scatters from a target material of atomic number 79. From the given data the estimated diameter of nuclei of the target material is (approximately) _____ m.

$\left[\dfrac{1}{4\pi\varepsilon_0}=9\times10^9\ \text{Nm}^2\text{/C}^2,\ e = 1.6\times10^{-19}\ \text{C}\right]$
A. $2.88 \times 10^{-14}$
B. $5.76 \times 10^{-14}$
C. $1.44 \times 10^{-13}$
D. $1.69 \times 10^{-12}$
Correct Answer: $5.76 \times 10^{-14}$ m

Explanation

In Rutherford scattering, the distance of closest approach corresponds to the point where the entire kinetic energy of the $\alpha$-particle is converted into electrostatic potential energy.

Energy of $\alpha$-particle:

$$ E = 7.9\ \text{MeV} = 7.9 \times 10^6 \times 1.6 \times 10^{-19} $$

$$ E = 1.264 \times 10^{-12}\ \text{J} $$

Charge of $\alpha$-particle $= +2e$ and charge of nucleus $= +Ze = 79e$.

Electrostatic potential energy at closest approach:

$$ \frac{1}{4\pi\varepsilon_0}\frac{(2e)(79e)}{r} = E $$

Substitute values:

$$ 9\times10^9 \times \frac{2\times79\times(1.6\times10^{-19})^2}{r} = 1.264\times10^{-12} $$

$$ \frac{9\times10^9 \times 158 \times 2.56\times10^{-38}}{r} = 1.264\times10^{-12} $$

$$ r = 2.88 \times 10^{-14}\ \text{m} $$

This gives the radius of the nucleus. Hence, diameter:

$$ D = 2r = 2 \times 2.88 \times 10^{-14} $$

$$ D = 5.76 \times 10^{-14}\ \text{m} $$

Therefore, the estimated diameter of the nucleus is $5.76 \times 10^{-14}$ m.

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