Question

The total kinetic energy carried by the products of the spontaneous fission of plutonium – 240 is typically about 180 MeV. Use this to argue there that reduction in Coulomb repulsion is the major impulse behind the process. Assume for simplicity that the two fragment nuclei are of equal Z.

Short answer

By fission, the binding energy would increase by $269.57\text{\hspace{0.17em}MeV}$ the Coulomb term, which contributes to the $180\text{\hspace{0.17em}MeV}$ kinetic energy of the daughter particle.

Step-by-step solution

Given data

Energy $=180\text{\hspace{0.17em}MeV}$.

Formula of Semi-empirical binding Energy

Semi-empirical binding energy is given by the expression: $\mathbf{BE}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{A}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}{\mathbf{A}}^{\mathbf{2}\mathbf{/}\mathbf{3}}\mathbf{-}{\mathbf{c}}_{\mathbf{3}}\frac{\mathbf{Z}\mathbf{(}\mathbf{Z}\mathbf{-}\mathbf{1}\mathbf{)}}{{\mathbf{A}}^{\mathbf{1}\mathbf{/}\mathbf{3}}}\mathbf{-}{\mathbf{c}}_{\mathbf{4}}\frac{{\mathbf{(}\mathbf{N}\mathbf{-}\mathbf{Z}\mathbf{)}}^{\mathbf{2}}}{\mathbf{A}}$ (1)

Where,${\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{15.8}\mathbf{MeV}\mathbf{,}{\mathbf{c}}_{\mathbf{2}}\mathbf{=}\mathbf{17.8}\mathbf{MeV}\mathbf{,}{\mathbf{c}}_{\mathbf{3}}\mathbf{=}\mathbf{0.71}\mathbf{MeV}\mathbf{,}\mathbf{\text{\hspace{0.17em}}}{\mathbf{c}}_{\mathbf{4}}\mathbf{=}\mathbf{23.7}\mathbf{MeV}$,$\mathbf{A}$is the mass number,$\mathbf{Z}$is the atomic number, and$\mathbf{N}$is the number of neutrons.

Check the Binding energy

The fission of plutonium- 240 into two particles with equal atomic numbers is:

${}_{84}^{240}\text{Pu}{\to }_{42}^{100}\text{Mo}{+}_{42}^{100}\text{Mo}+40\mathrm{n}$

The semi-empirical binding energy formula is,

$\text{BE}={\text{c}}_1\mathrm{A}-{\mathrm{c}}_2{\mathrm{A}}^{2/3}-{\mathrm{c}}_3\frac{\mathrm{Z}(\mathrm{Z}-1)}{\mathrm{A}}-{\mathrm{c}}_4\frac{{(\mathrm{Z}-1)}^2}{\mathrm{A}}$

And

$\mathrm{Z}=84,\mathrm{A}=240,\mathrm{N}=240-84=156$

The third term contribution to plutonium- 240 is,

${\mathrm{BE}}_3=-{\mathrm{c}}_3\frac{\mathrm{Z}(\mathrm{Z}-1)}{{\mathrm{A}}^{1/3}}$

Substitute values in the above equation, and we get,

$\begin{array}{c}{\mathrm{BE}}_3=-\left(0.71\text{\hspace{0.17em}MeV}\right)\frac{\left(84\right)(84-1)}{{\left(240\right)}^{1/3}}\\ =-796.55\text{\hspace{0.17em}MeV}\end{array}$

$\mathrm{The}\mathrm{third}\mathrm{term}\mathrm{contribution}\mathrm{to}\mathrm{molybdenum}-100\mathrm{is},$

${\mathrm{BE}}_3=-{\mathrm{c}}_3\frac{\mathrm{Z}(\mathrm{Z}-1)}{{\mathrm{A}}^{1/3}}$

And

$\mathrm{Z}=42,\mathrm{A}=100,\mathrm{N}=100-42=58$

Substitute values in the above equation, and we get,

$\begin{array}{c}{\mathrm{BE}}_3=-\left(0.71\text{\hspace{0.17em}MeV}\right)\frac{\left(42\right)(42-1)}{{\left(100\right)}^{1/3}}\\ =-263.49\text{\hspace{0.17em}MeV}\end{array}$

So the difference in the binding energy of the parent particle and the daughter particles is:

$\begin{array}{l}=-796.55\text{\hspace{0.17em}MeV}-2(-263.49\text{\hspace{0.17em}MeV})\\ =-269.57\text{\hspace{0.17em}MeV}\end{array}$

By fission, the binding energy would increase by $269.57\text{\hspace{0.17em}MeV}$ by the Coulomb term, which contributes to the$180\text{\hspace{0.17em}MeV}$ kinetic energy of the daughter particle.