Chapter 46, Problem 14SP

To determine

The binding energy of carbon-12 $\left({}^{12}\text{C with Z=6}\right)$ whose atomic mass is $12.\text{000000 u}$.

Answer to Problem 14SP

Solution:

$92.16\text{ MeV}$

Explanation of Solution

Given data:

The mass of the carbon atom is $12.\text{000000 u}$.

The atomic mass number of the carbon atom is $12$.

Formula used:

The expression for the atomic mass number of an atom is written as

$A=Z+N$

Here, $Z$ is the atomic number and $N$ is the number of neutrons.

The expression for the binding energy of the nucleus is written as

$\text{BE}\approx \left(Z{M}_{\text{H}}+N{M}_n-M_A\right)c^2$

Here, $M_{\text{H}}$ is the mass of hydrogen, $M_n$ is the mass of neutron, $M_A$ is the atomic mass, and $c$ is the speed of light.

Explanation:

Recall the expression for the atomic mass number of an atom:

$A=Z+N$

Rearrange the expression for $N$

$N=A-Z$

Substitute $12$ for $A$ and $6$ for $Z$

$\begin{array}{c}N=12-6\\ =6\end{array}$

This is the number of neutrons present in carbon-12.

Write the expression for the binding energy of carbon-12:

$\text{BE}\approx \left(Z{M}_{\text{H}}+N{M}_n-M_{\text{C}}\right)c^2$

Here, $M_{\text{C}}$ is the mass of the carbon atom.

Substitute $6$ for $Z$, $6$ for $N$, $1.007825\text{ u}$ for $M_{\text{H}}$, $12.\text{000000 u}$ for $M_{\text{C}}$, and $1.008665\text{ u}$ for $M_N$

$\begin{array}{c}\text{BE}\approx \left[6\left(1.007825\text{ u}\right)+6\left(1.008665\text{ u}\right)-12.\text{000000 u}\right]c^2\\ =\left(0.09894\text{ u}\right)\left(\frac{931.494095\text{ MeV}}{c^2}\right)c^2\\ =92.16\text{ MeV}\end{array}$

Conclusion:

Hence, the binding energy of carbon-12 is $92.16\text{ MeV}$.