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Energy $=180\text{鈥塎别痴}$.

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

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

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

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

The semi-empirical binding energy formula is,

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

And

$\mathrm{Z}=84,\mathrm{A}=240,\text{鈥�}\mathrm{N}=240鈭�84=156$

The third term contribution to plutonium- 240 is,

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

Substitute values in the above equation, and we get,

$\begin{array}{c}{\mathrm{BE}}_3=鈭�\left(0.71\text{鈥塎别痴}\right)\frac{\left(84\right)(84鈭�1)}{{\left(240\right)}^{1/3}}\\ =鈭�796.55\text{鈥塎别痴}\end{array}$

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

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

And

$\mathrm{Z}=42,\mathrm{A}=100,\text{鈥�}\mathrm{N}=100鈭�42=58$

Substitute values in the above equation, and we get,

$\begin{array}{c}{\mathrm{BE}}_3=鈭�\left(0.71\text{鈥塎别痴}\right)\frac{\left(42\right)(42鈭�1)}{{\left(100\right)}^{1/3}}\\ =鈭�263.49\text{鈥塎别痴}\end{array}$

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

$\begin{array}{l}=鈭�796.55\text{鈥塎别痴}鈭�2(鈭�263.49\text{鈥塎别痴})\\ =鈭�269.57\text{鈥塎别痴}\end{array}$

By fission, the binding energy would increase by $269.57\text{鈥塎别痴}$ by the Coulomb term, which contributes to the$180\text{鈥塎别痴}$ kinetic energy of the daughter particle.