Question

What does the semiemperical binding energy formula predict forthe mass of a carbon- 12 atom? Comment on your answer.

Short answer

The resultant answeris $12.005542u$.

Step-by-step solution

Given data

Carbon 12-atom is the given data.

Formula of Semi empirical binding Energy

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

Where,$c_1=15.8MeV,c_2=17.8MeV,c_3=0.71MeV,\text{\hspace{0.17em}}{c}_4=23.7MeV$.

$A\to$Mass Number,$Z\to$atomic Number and$N\to$number of neutrons.

Calculate the semi empirical binding energy

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

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

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

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.

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

Substitute the values

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

Rearrange the above equation for$M_C$:

$M_C=Z{M}_H+N{m}_u-\frac{BE}{c^2}$

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

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

This is slightly larger than its real mass.