Question

The electrostatic energy of \(Z\) protons uniformly distributed throughout a spherical nucleus of radius \(R\) is given by $$ E=\frac{3}{5} \frac{Z(Z-1) e^{2}}{4 \pi \varepsilon_{0} R} $$ The measured masses of the neutron, \({ }_{1}^{1} \mathrm{H},{ }_{7}^{15} \mathrm{~N}\) and \({ }_{8}^{15} \mathrm{O}\) are \(1.008665 \mathrm{u}, 1.007825 \mathrm{u}\), \(15.000109 \mathrm{u}\) and \(15.003065 \mathrm{u}\), respectively. Given that the radii of both the \({ }_{7}^{15} \mathrm{~N}\) and \({ }_{8}^{15} \mathrm{O}\) nuclei are same, \(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2}\left(c\right.\) is the speed of light) and \(e^{2} /\left(4 \pi \varepsilon_{0}\right)=1.44 \mathrm{MeV} \mathrm{fm}\). Assuming that the difference between the binding energies of \({ }_{7}^{15} \mathrm{~N}\) and \({ }_{8}^{15} \mathrm{O}\) is purely due to the electrostatic energy, the radius of either of the nuclei is \(\left(1 \mathrm{fm}=10^{-15} \mathrm{~m}\right)\) (A) \(2.85 \mathrm{fm}\) (B) \(3.03 \mathrm{fm}\) (C) \(3.42 \mathrm{fm}\) (D) \(3.80 \mathrm{fm}\)

Short answer

3.03 fm

Step-by-step solution

Calculate the Binding Energy Difference

First, use the given mass values to find the binding energy difference between _{7}^{15}N and _{8}^{15}O. Convert the mass defect to energy using the conversion factor 1 u = 931.5 MeV/c^2. The binding energy difference (in MeV) is the difference between the converted mass defects of the two nuclei.

Determine the Charge Difference

Calculate the difference in charge, which corresponds to the difference in the number of protons in the two nuclei. Since _{7}^{15}N has 7 protons and _{8}^{15}O has 8 protons, the charge difference is 1e where e is the elementary charge.

Calculate the Electrostatic Energy Difference

The electrostatic energy difference between the two nuclei can be found using the formula in the exercise and the charge difference: \[ \Delta E=\frac{3}{5}\frac{e^2}{4\pi\varepsilon_0}\left(\frac{Z_{O}(Z_{O}-1)}{R_{O}}-\frac{Z_{N}(Z_{N}-1)}{R_{N}}\right) \]. Since the radii of both nuclei are the same, we denote them as R.

Isolate the Radius (R)

To find the radius, equate the electrostatic energy difference \(\Delta E\) to the binding energy difference obtained in Step 1 and solve for R.

Substitute Known Values and Solve for Radius

Now substitute the given values \(e^2/(4\pi\varepsilon_0) = 1.44\) MeV fm, the calculated binding energy difference, and the values for Z into the formula and solve for the radius R.

Key concepts

Binding Energy Difference

The concept of binding energy is fundamental to understanding nuclear stability and reactions. In simple terms, binding energy represents the energy required to disassemble a nucleus into its component protons and neutrons. The greater the binding energy, the more stable the nucleus.

The binding energy difference is particularly interesting when comparing isotopes or isotones. In the given exercise, students are asked to compute this difference between nitrogen-15 and oxygen-15. This is done by converting the mass defects of both nuclei into energy, which indicates how much more or less energy is stored in the bonds of the respective nuclei.

To visualize this, imagine the nucleus as a tightly packed cluster of marbles (protons and neutrons). The binding energy is like the strength of glue holding these marbles together. If we compare two clusters — for example, one with a slightly heavier marble representing an additional proton as in oxygen-15 compared to nitrogen-15 — the difference in the amount of glue needed highlights the binding energy difference between them.

Nuclear Radius Calculation

Calculating the nuclear radius is akin to estimating the size of a tiny, yet incredibly dense sphere of nuclear matter. This is not observable through ordinary microscopes; therefore, physicists have developed equations that relate a nucleus's radius to its properties.

In the given problem, an equation correlating electrostatic energy and nuclear radius serves as the basis for our calculation. Once the binding energy difference, which partially arises from electrostatic interactions between protons, is known, we can rearrange that equation to isolate and solve for the radius (R).

The calculation assumes that the radius of nuclei with the same mass number is identical, which highlights a key fact in nuclear physics: the nucleus is approximately spherical, and its size does not grow significantly even as we add more protons or neutrons, up to a certain limit. This is because the nuclear force pulling these particles together is much stronger than the electrostatic force trying to push the protons apart, at least within the size limits of stable nuclei.

Proton Distribution in Nucleus

The distribution of protons within the nucleus has a direct impact on the nuclear electrostatic energy. In a simplified model, we often assume protons are uniformly distributed throughout a spherical nucleus. This model helps us understand electrostatic repulsion forces between positively charged protons.

Within the nucleus, the proton distribution affects the calculation of electrostatic energy. In the exercise, the difference in proton number between nitrogen-15 and oxygen-15 alters the electrostatic energy, which, when calculated, allows us to derive other properties such as the nuclear radius.

It’s important to understand that this uniform distribution is an approximation. In reality, the quantum mechanical nature of protons leads to a probability distribution rather than a precise location. However, for many basic calculations and to a first approximation, uniform distribution is a useful and sufficiently accurate model.