The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius

Q: Why do we use the mass of a hydrogen atom $_1^1\text{H}$ instead of just a proton?

Using the mass of a hydrogen atom automatically accounts for the mass of the electrons in the neutral atom. Since the given masses are atomic masses (not bare nuclear masses), the electron masses cancel out perfectly when we use the hydrogen atom mass.

Q: Why is the electrostatic energy proportional to $Z(Z-1)$ instead of $Z^2$?

A proton does not repel itself. The electrostatic energy comes from the interaction between *pairs* of protons. The number of unique pairs you can form from $Z$ protons is $Z(Z-1)/2$.

Q: What are mirror nuclei?

$^{15}\text{N}$ and $^{15}\text{O}$ are an example of mirror nuclei. They have the same mass number $A$, but the number of protons in one equals the number of neutrons in the other, and vice versa. Their nuclear strong force interactions are nearly identical, so any difference in binding energy is almost entirely due to electrostatic repulsion.

Q: Why did we convert the mass defect to MeV?

The electrostatic constant is given in terms of $\text{MeV fm}$. To equate the mass defect to the electrostatic energy, both must be in the same energy units. We use the conversion factor $1 \text{ u} = 931.5 \text{ MeV/c}^2$.

Q: Is the radius of a nucleus really the same for different isotopes?

The radius of a nucleus is approximately given by $R = R_0 A^{1/3}$. Since both $^{15}\text{N}$ and $^{15}\text{O}$ have the same mass number $A=15$, their radii are expected to be very similar, which is why the problem assumes they are the same.

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, $_1^1	ext{H}$, $_7^{15}	ext{N}$ and $_8^{15}	ext{O}$ are $1.008665 	ext{ u}$, $1.007825 	ext{ u}$, $15.000109 	ext{ u}$ and $15.003065 	ext{ u}$, respectively. Given that the radii of both the $_7^{15}	ext{N}$ and $_8^{15}	ext{O}$ nuclei are same, $1 	ext{ u} = 931.5 	ext{ MeV/c}^2$ ($c$ is the speed of light) and $e^2/(4\pi\varepsilon_0) = 1.44 	ext{ MeV fm}$. Assuming that the difference between the binding energies of $_7^{15}	ext{N}$ and $_8^{15}	ext{O}$ is purely due to the electrostatic energy, the radius of either of the nuclei is ($1 	ext{ fm} = 10^{-15} 	ext{ m}$)

Options:

(A) 2.85 fm

(B) 3.03 fm

(C) 3.42 fm

(D) 3.80 fm

Sigma Percentile · Accepted Answer

### Analyzing the Setup
The problem presents us with a fascinating scenario involving two mirror nuclei: Nitrogen-15 ($^{15}_7	ext{N}$) and Oxygen-15 ($^{15}_8	ext{O}$). Mirror nuclei are pairs of nuclei where the number of protons in one equals the number of neutrons in the other. Because the strong nuclear force is charge-independent, the nuclear interactions in both nuclei are nearly identical. Therefore, any difference in their total binding energies arises almost entirely from the electrostatic repulsion between the protons.

We are given the formula for the electrostatic energy of a uniformly charged spherical nucleus:
$$E = \frac{3}{5} \frac{Z(Z-1)e^2}{4\pi\varepsilon_0 R}$$
Notice the term $Z(Z-1)$. This is because a proton does not repel itself; the electrostatic energy comes from the interaction between pairs of protons, and the number of pairs is proportional to $Z(Z-1)$.

### The Master Equation
The binding energy ($B.E.$) of a nucleus is the energy required to disassemble it into its constituent protons and neutrons. It is calculated from the mass defect:
$$B.E. = [Z M_H + (A-Z) M_n - M_{	ext{nucleus}}]c^2$$
Here, we use the mass of a hydrogen atom ($M_H$) instead of a bare proton to automatically account for the mass of the electrons in the neutral atomic masses given in the problem.

Let's calculate the difference in binding energies between $^{15}_7	ext{N}$ and $^{15}_8	ext{O}$:
$$\Delta B.E. = B.E.(^{15}	ext{N}) - B.E.(^{15}	ext{O})$$
$$\Delta B.E. = [7 M_H + 8 M_n - M_N]c^2 - [8 M_H + 7 M_n - M_O]c^2$$
$$\Delta B.E. = [M_n - M_H + M_O - M_N]c^2$$
This elegant cancellation leaves us with a very simple expression!

### Raw Setup and Substitution
Now, we carefully substitute the given mass values:
$$\Delta B.E. = [1.008665 - 1.007825 + 15.003065 - 15.000109] 	ext{ u} 	imes c^2$$
$$\Delta B.E. = [0.000840 + 0.002956] 	ext{ u} 	imes c^2 = 0.003796 	ext{ u} 	imes c^2$$
To convert this mass defect into energy, we use the conversion factor $1 	ext{ u} = 931.5 	ext{ MeV/c}^2$:
$$\Delta B.E. = 0.003796 	imes 931.5 	ext{ MeV} \approx 3.536 	ext{ MeV}$$

### Final Calculation
This difference in binding energy is purely due to the difference in electrostatic energy ($\Delta E$):
$$\Delta E = E_O - E_N = \frac{3}{5} \frac{e^2}{4\pi\varepsilon_0 R} [Z_O(Z_O-1) - Z_N(Z_N-1)]$$
Substitute $Z_O = 8$ and $Z_N = 7$:
$$\Delta E = \frac{3}{5} \frac{1.44}{R} [8(7) - 7(6)] = \frac{3}{5} \frac{1.44}{R} [56 - 42] = \frac{3}{5} \frac{1.44 	imes 14}{R}$$
Equating the two energy differences:
$$3.536 = \frac{3}{5} \frac{1.44 	imes 14}{R}$$
$$R = \frac{3 	imes 1.44 	imes 14}{5 	imes 3.536} = \frac{60.48}{17.68} \approx 3.42 	ext{ fm}$$
**The radius of either of the nuclei is 3.42 fm.**

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Visualized Solution

$Binding Energy Formula$

$Difference in Binding Energies$

$Substituting Mass Values$

$Calculating Δ B.E.$

$Difference in Electrostatic Energies$

$Substituting Z and Constants$

$Equating and Solving for R$

$Final Answer$

The Sigma Insight: Nucleus and Nuclear Reaction

Analyzing the Setup The problem presents us with a fascinating scenario involving two mirror nuclei

The Master Equation The binding energy ( $B . E .$ ) of a nucleus is the energy required to disassemble it into its constituent protons and neutrons

Raw Setup and Substitution

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Binding Energy Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Difference in Binding Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Mass Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating ΔB.E.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Difference in Electrostatic Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Z and Constants

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating and Solving for R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer