JEE Challenger
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PhyMedAdv 2016 • Paper 2

Electrostatic Energy and Binding Energy Radius Calculation of Nuclei

For a spherical nucleus of radius RR, the electrostatic energy resulting from ZZ uniformly distributed protons is given by E=35Z(Z−1)e24πε0RE = \frac{3}{5} \frac{Z(Z - 1)e^2}{4\pi\varepsilon_0 R} The measured masses of a neutron, 11H^1_1\text{H}, 715N^{15}_7\text{N}, and 815O^{15}_8\text{O} are 1.008665 u1.008665\text{ u}, 1.007825 u1.007825\text{ u}, 15.000109 u15.000109\text{ u}, and 15.003065 u15.003065\text{ u}, respectively. Given that both 715N^{15}_7\text{N} and 815O^{15}_8\text{O} nuclei possess the same radius, where 1 u=931.5 MeV/c21\text{ u} = 931.5\text{ MeV}/c^2 (cc denotes the speed of light) and e2/(4πε0)=1.44 MeV fme^2 / (4\pi\varepsilon_0) = 1.44\text{ MeV fm}. Assuming that the difference in binding energies between 715N^{15}_7\text{N} and 815O^{15}_8\text{O} is solely due to electrostatic energy, find the radius of either nucleus (1 fm=10−15 m1\text{ fm} = 10^{-15}\text{ m}).

Options

A

2.85 fm2.85\text{ fm}

B

3.03 fm3.03\text{ fm}

C

3.42 fm3.42\text{ fm}

Correct
D

3.80 fm3.80\text{ fm}

Topics & Concepts

NucleiNuclear ReactionsElectric Potential

Step-by-Step Solution

To find the radius RR of the nuclei, we use the principle that the difference in binding energy between the mirror nuclei 715N^{15}_7\text{N} and 815O^{15}_8\text{O} is entirely due to the difference in their electrostatic potential energies.

1. Calculation of Binding Energy Difference

The binding energy BEBE of a nucleus with mass number AA and atomic number ZZ is given in terms of atomic masses by: BE(Z,A)=[Z m(11H)+(A−Z) mn−M(Z,A)]c2BE(Z, A) = \left[ Z\, m\left(^1_1\text{H}\right) + (A - Z)\, m_n - M(Z, A) \right] c^2

For 715N^{15}_7\text{N} (Z=7Z = 7, N=8N = 8): BE(715N)=[7 m(11H)+8 mn−M(715N)]c2BE\left(^{15}_7\text{N}\right) = \left[ 7\, m\left(^1_1\text{H}\right) + 8\, m_n - M\left(^{15}_7\text{N}\right) \right] c^2

For 815O^{15}_8\text{O} (Z=8Z = 8, N=7N = 7): BE(815O)=[8 m(11H)+7 mn−M(815O)]c2BE\left(^{15}_8\text{O}\right) = \left[ 8\, m\left(^1_1\text{H}\right) + 7\, m_n - M\left(^{15}_8\text{O}\right) \right] c^2

The difference in binding energies ΔBE=BE(715N)−BE(815O)\Delta BE = BE\left(^{15}_7\text{N}\right) - BE\left(^{15}_8\text{O}\right) is: ΔBE=[(mn−m(11H))+(M(815O)−M(715N))]c2\Delta BE = \left[ (m_n - m\left(^1_1\text{H}\right)) + \left( M\left(^{15}_8\text{O}\right) - M\left(^{15}_7\text{N}\right) \right) \right] c^2

Substitute the given mass values:

  • mn−m(11H)=1.008665 u−1.007825 u=0.000840 um_n - m\left(^1_1\text{H}\right) = 1.008665\text{ u} - 1.007825\text{ u} = 0.000840\text{ u}
  • M(815O)−M(715N)=15.003065 u−15.000109 u=0.002956 uM\left(^{15}_8\text{O}\right) - M\left(^{15}_7\text{N}\right) = 15.003065\text{ u} - 15.000109\text{ u} = 0.002956\text{ u}

Therefore, the total mass difference equivalent is: Δm=0.000840 u+0.002956 u=0.003796 u\Delta m = 0.000840\text{ u} + 0.002956\text{ u} = 0.003796\text{ u}

Converting this mass difference into energy using 1 u=931.5 MeV/c21\text{ u} = 931.5\text{ MeV}/c^2: ΔBE=0.003796×931.5 MeV=3.535974 MeV\Delta BE = 0.003796 \times 931.5\text{ MeV} = 3.535974\text{ MeV}

2. Calculation of Electrostatic Energy Difference

The electrostatic energy of a uniformly charged sphere with ZZ protons is given as: E=35Z(Z−1)e24πε0RE = \frac{3}{5} \frac{Z(Z - 1)e^2}{4\pi\varepsilon_0 R}

The difference in electrostatic energy between 815O^{15}_8\text{O} (Z=8Z = 8) and 715N^{15}_7\text{N} (Z=7Z = 7) is: ΔEe=Ee(815O)−Ee(715N)=35e24πε0R[8(8−1)−7(7−1)]\Delta E_e = E_e\left(^{15}_8\text{O}\right) - E_e\left(^{15}_7\text{N}\right) = \frac{3}{5} \frac{e^2}{4\pi\varepsilon_0 R} \left[ 8(8 - 1) - 7(7 - 1) \right]

ΔEe=35e24πε0R[56−42]=35×e24πε0R×14\Delta E_e = \frac{3}{5} \frac{e^2}{4\pi\varepsilon_0 R} \left[ 56 - 42 \right] = \frac{3}{5} \times \frac{e^2}{4\pi\varepsilon_0 R} \times 14

Given that e24πε0=1.44 MeV fm\frac{e^2}{4\pi\varepsilon_0} = 1.44\text{ MeV fm}: ΔEe=3×14×1.445R=12.096R MeV⋅fm\Delta E_e = \frac{3 \times 14 \times 1.44}{5 R} = \frac{12.096}{R}\text{ MeV}\cdot\text{fm}

3. Equating ΔBE\Delta BE and ΔEe\Delta E_e

Equating the difference in binding energy to the electrostatic energy difference: ΔBE=ΔEe\Delta BE = \Delta E_e 3.535974 MeV=12.096R MeV⋅fm3.535974\text{ MeV} = \frac{12.096}{R}\text{ MeV}\cdot\text{fm}

Solving for RR: R=12.0963.535974 fm≈3.4208 fm≈3.42 fmR = \frac{12.096}{3.535974}\text{ fm} \approx 3.4208\text{ fm} \approx 3.42\text{ fm}

Conclusion

The radius of either nucleus is 3.42 fm3.42\text{ fm}.

Correct Option: (C)

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