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Let's analyze the problem step by step:

Given:

Electrostatic energy: $E = \frac{3}{5} \frac{Z (Z - 1) e^{2}}{4 π ϵ_{0} R}$
Masses: neutron ( $m_{n}$ ) = 1.008665 u, $_{1}^{1} H$ ( $m_{H}$ ) = 1.007825 u, $_{7}^{15} N$ = 15.000109 u, $_{8}^{15} O$ = 15.003065 u.
1 u = 931.5 MeV/ $c^{2}$
$e^{2} / (4 π ϵ_{0}) = 1.44$ MeV·fm
Radii of $_{7}^{15} N$ and $_{8}^{15} O$ are the same.
The difference in binding energies (BE) is purely due to electrostatic energy.

We need to find the radius $R$ .

Step 1: Binding Energy (BE) Calculation
The binding energy for a nucleus with $Z$ protons and $N$ neutrons is:

B E = [Z m_{H} + N m_{n} - M_{nucleus}] c^{2}

For $_{7}^{15} N$ : $Z = 7, N = 8$

B E_{N} = [7 \times m_{H} + 8 \times m_{n} - M_{N}] c^{2}

For $_{8}^{15} O$ : $Z = 8, N = 7$

B E_{O} = [8 \times m_{H} + 7 \times m_{n} - M_{O}] c^{2}

Step 2: Difference in Binding Energies

Δ B E = B E_{O} - B E_{N}

Substitute:

Δ B E = ([8 m_{H} + 7 m_{n} - M_{O}] - [7 m_{H} + 8 m_{n} - M_{N}]) c^{2}

= [(8 m_{H} - 7 m_{H}) + (7 m_{n} - 8 m_{n}) - (M_{O} - M_{N})] c^{2}

= [m_{H} - m_{n} - (M_{O} - M_{N})] c^{2}

Now plug in the masses (in u):
$m_{H} = 1.007825$ u, $m_{n} = 1.008665$ u, $M_{O} = 15.003065$ u, $M_{N} = 15.000109$ u.
So,

M_{O} - M_{N} = 15.003065 - 15.000109 = 0.002956 u

m_{H} - m_{n} = 1.007825 - 1.008665 = - 0.000840 u

Thus,

Δ B E = [- 0.000840 - 0.002956] \times 931.5 MeV

= [- 0.003796] \times 931.5 MeV

= - 3.535 MeV

So, $B E_{O} - B E_{N} = - 3.535$ MeV, meaning $B E_{O} < B E_{N}$ by 3.535 MeV.

Step 3: Electrostatic Energy Difference
The electrostatic energy for a nucleus with proton number $Z$ is:

E_{es} = \frac{3}{5} \frac{Z (Z - 1) e^{2}}{4 π ϵ_{0} R}

For $_{7}^{15} N$ : $Z = 7$

E_{es,N} = \frac{3}{5} \frac{7 \cdot 6 \cdot e^{2}}{4 π ϵ_{0} R} = \frac{3}{5} \frac{42 \cdot e^{2}}{4 π ϵ_{0} R}

For $_{8}^{15} O$ : $Z = 8$

E_{es,O} = \frac{3}{5} \frac{8 \cdot 7 \cdot e^{2}}{4 π ϵ_{0} R} = \frac{3}{5} \frac{56 \cdot e^{2}}{4 π ϵ_{0} R}

The difference in electrostatic energy:

Δ E_{es} = E_{es,O} - E_{es,N} = \frac{3}{5} \frac{e^{2}}{4 π ϵ_{0} R} (56 - 42) = \frac{3}{5} \frac{14 \cdot e^{2}}{4 π ϵ_{0} R}

Given that the difference in binding energy is purely due to electrostatic energy, we have:

Δ B E = Δ E_{es}

But note: Higher electrostatic energy reduces binding energy. So,

B E_{O} - B E_{N} = - (E_{es,O} - E_{es,N}) = - Δ E_{es}

Therefore:

- Δ E_{es} = - 3.535 MeV

Δ E_{es} = 3.535 MeV

So,

\frac{3}{5} \frac{14 \cdot e^{2}}{4 π ϵ_{0} R} = 3.535

Step 4: Solve for $R$
Given $e^{2} / (4 π ϵ_{0}) = 1.44$ MeV·fm.
So,

\frac{3}{5} \cdot \frac{14 \cdot 1.44}{R} = 3.535

Compute:

\frac{3}{5} \cdot 14 = \frac{42}{5} = 8.4

So,

8.4 \cdot \frac{1.44}{R} = 3.535

\frac{12.096}{R} = 3.535

R = \frac{12.096}{3.535} \approx 3.421 fm

Thus, the radius is approximately 3.42 fm.

Final Answer:

3.42 fm