91影视

Carbon 12-atom is the given data.

Semi empirical binding Energy is given by the expression, $BE=c_{1}A鈭�c_2{A}^{2/3}鈭�c_3\frac{Z(Z鈭�1)}{A^{1/3}}鈭�c_4\frac{{(N鈭�Z)}^2}{A}$.

Where,$c_1=15.8MeV,\text{鈥�}{c}_2=17.8MeV,\text{鈥�}{c}_3=0.71MeV,\text{鈥嬧赌�}{c}_4=23.7MeV$.

$A鈫�$Mass Number,$Z鈫�$atomic Number and$N鈫�$number of neutrons.

The binding energy calculated by the semi empirical binding energy formula for a carbon- 12 atom is:

Has$Z=6$ and$A=12$ .

$\begin{array}{c}N=A鈭�Z\\ N=鈭�12鈭�6\\ N=6\\ BE=c_{1}A鈭�c_2{A}^{2/3}鈭�c_3\frac{Z(Z鈭�1)}{A^{1/3}}鈭�c_4\frac{{(N鈭�Z)}^2}{A}\end{array}$

Substitute$15.8\text{MeV}$ for $c_1,17.8\text{MeV}$for$c_2,0.71\text{MeV}$for$c_3,43$ for$Z$, $98for\text{A}$and $55forN$ in above equation.

$\begin{array}{l}BE=15.8\left(12\right)鈭�17.8{\left(12\right)}^{2/3}鈭�0.71\frac{6(6鈭�1)}{{\left(12\right)}^{1/3}}鈭�23.7\frac{{(6鈭�6)}^2}{12}\\ BE=87.00\text{MeV}\end{array}$

By the other model, we have.$BE=(Z{M}_H+N{m}_u鈭�M_c)c^2$

Rearrange the above equation for$M_C$:

$M_C=Z{M}_H+N{m}_u鈭�\frac{BE}{c^2}$

Substitute $1.007825u{\text{M}}_H,1.008665u$ for $m_u$ and $87.00\text{MeV}$ for BE in above equation.

$\begin{array}{l}{M}_C=6\left(1.007825u\right)+6\left(1.008665u\right)鈭�\frac{\left(87.00\text{MeV}\right)}{(931.5\text{MeV}/u)}\\ M_C=12.005542u\end{array}$

This is slightly larger than its real mass.