Question

Use the graph of binding energy to estimate the total energy released if three ${}^4\text{He}$nuclei fuse together to form a${}^{12}\text{C}$ nucleus.

Short answer

The amount of energy released is$6.0\text{MeV}$.

Step-by-step solution

Step.1.

From the figure 42.6, plot binding energy per nucleon verse mass number given in the text book, the binding energy per nucleon for ${}^4\text{He}$nucleus approximately is $7.2\text{MeV}$and the binding energy per nucleon for ${}^{12}\text{C}$nucleus is $7.7\text{MeV}$.

The nucleus${}^4\text{He}$has 4 nucleons, so the total binding energy of one ${}^4\text{He}$nucleus is,

$\begin{array}{l}{\left(\Delta E\right)}_{{}^4\text{He}}=\left(4\right)\left(7.2\text{MeV}\right)\\ =28.8\text{MeV}\end{array}$

The total binding energy of three${}^4\text{He}$nuclei is,

$\begin{array}{l}\Delta E_1=\left(3\right)\left(28.8\text{MeV}\right)\\ =86.4\text{MeV}\end{array}$

The nucleus ${}^{12}\text{C}$has 12 nucleons, so the total binding energy of ${}^{12}\text{C}$nucleus is,

$\begin{array}{l}\Delta E_2=\left(12\right)\left(7.7\text{MeV}\right)\\ =92.4\text{MeV}\end{array}$

 Step.2.

The amount of energy released in the nuclear reaction which involves three ${}^4\text{He}$nuclei fuse together and form a${}^{12}\text{C}$nucleus is,

$\begin{array}{l}\Delta E=E_2-E_1\\ =92.4\text{MeV}-86.4\text{MeV}\\ =6.0\text{MeV}\end{array}$

Therefore, the amount of energy released is $6.0\text{MeV}$.