Question

The electric potential energy of a uniform sphere of charge qand radius ris given by $\mathit{U}\mathbf{=}\frac{\mathbf{3}{\mathbf{q}}^{\mathbf{2}}}{\mathbf{20}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{r}}$(a) Does the energy represents a tendency for the sphere to bind together or blow apart? The nuclide ${}^{\mathbf{239}}\mathbf{Pu}$ is spherical with radius 6.64fm . For this nuclide, what are (b) the electric potential energy Uaccording to the equation, (c) the electric potential energy per proton, and (d) the electric potential energy per nucleon? The binding energy per nucleon is $\mathbf{7}\mathbf{.}\mathbf{56}\mathbf{}\mathbf{MeV}$. (e) Why is the nuclide bound so well when the answers to (c) and (d) are large and positive?

Short answer

1. Yes, the energy represents the tendency for the sphere to bind together.
2. The electric potential energy of the nuclide ${}^{239}\mathrm{Pu}$according to the equation is 1.15 GeV .
3. The electric potential energy per proton is 12.2 MeV/proton .
4. The electric potential energy per nucleon is 4.81 MeV/nucleon .
5. The nuclide is bound so well because of the strong force.

Step-by-step solution

Given data

The given equation for the electric potential energy of a uniform sphere of charge and radius can be given as follows: $\mathrm{U}=\frac{3{\mathrm{q}}^2}{20{\mathrm{\pi \epsilon }}_0\mathrm{r}}$

The radius of the nuclide ${}^{239}\mathrm{Pu},\mathrm{r}=6.64\mathrm{fm}$,

Binding energy per nucleon, B.E. = 7.56 MeV

Understanding the concept of energy  

Strong binding is represented by the positive potential energy of a system. Thus, a strong force gives high binding energy for the given nuclide. The binding energy of the system can be defined by the amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Thus, using the given data and the formula, we can solve the basic case of energy considering the proton and nucleon number of the nuclide.

Formula:

The given equation for the electric potential energy of a uniform sphere of charge and radius can be given as follows: $\mathrm{U}=\frac{3{\mathrm{q}}^2}{20{\mathrm{\pi \epsilon }}_0\mathrm{r}}........\left(1\right)$

a) Calculation of the tendency of binding

Since U > 0 , the energy represents a tendency for the sphere to blow apart.

Hence, the energy represents the tendency for the sphere to bind together.

b) Calculation of the electric potential energy

For the given nuclide ${}^{239}\mathrm{Pu}$, q = 94e and r = 6.64 fm , we can get the electric potential energy of the nuclide ${}^{239}\mathrm{Pu}$according to the given equation (1) as follows:

⁹N.m²/C²56.64×10-15m1eV1.6×10-19J=1.15×109eV=1.15GeV

Hence, the value of the energy is 1.15 GeV .

c) Calculation of the electric potential energy per proton

Using the above energy value and proton number as Z = 94 , we can get the required electric potential energy per proton of the nuclide as follows:

$$\begin{gathered} {\mathrm{U}}_{\mathrm{proton}}=1.15\mathrm{GeV}/94 \\ =12.2\mathrm{MeV}/\mathrm{proton} \end{gathered}$$

Hence, the value of the energy is 12.2MeV/proton .

d) Calculation of the electric potential energy per nucleon

Using the above energy value and proton number as A = 239 , we can get the required electric potential energy per proton of the nuclide as follows:

$$\begin{gathered} {\mathrm{U}}_{\mathrm{proton}}=1.15\mathrm{GeV}/239 \\ =4.81\mathrm{MeV}/\mathrm{nucleon} \end{gathered}$$

Hence, the value of the energy is 4.81 MeV/nucleon .

e) Calculation of the reason of binding of the nuclide

The strong force that binds the nucleus is very strong. Thus, the nuclide’s binding is directly proportional to the large and positive answers of part (c) and (d).