A carbon-12 atom has a mass defect of 0.09564 amu.

Which setup is used to calculate nuclear binding energy?

A. [tex]\(0.09564 \, \text{amu} \times \left(1.6606 \times 10^{-27} \, \text{kg} / \text{amu}\right) \times \left(3.0 \times 10^8 \, \text{m/s}\right)^2\)[/tex]

B. [tex]\(0.09564 \, \text{amu} \times 1 \, \text{amu} / \left(1.6606 \times 10^{-27} \, \text{kg}\right) \times \left(3.0 \times 10^8 \, \text{m/s}\right)^2\)[/tex]

C. [tex]\(0.09564 \, \text{amu} \times \left(3.0 \times 10^8 \, \text{m/s}\right)^2\)[/tex]

D. [tex]\(0.09564 \, \text{amu} \times \left(1.6606 \times 10^{-27} \, \text{kg} / \text{amu}\right) \times \left(3.0 \times 10^8 \, \text{m/s}\right)\)[/tex]



Answer :

To calculate the nuclear binding energy, we use Einstein's mass-energy equivalence principle given by the formula [tex]\(E = \Delta m \cdot c^2\)[/tex], where [tex]\(E\)[/tex] is the energy, [tex]\(\Delta m\)[/tex] is the mass defect, and [tex]\(c\)[/tex] is the speed of light.

Given:

1. The mass defect [tex]\(\Delta m = 0.09564 \, \text{amu}\)[/tex]
2. Conversion factor [tex]\(1 \, \text{amu} = 1.6606 \times 10^{-27} \, \text{kg}\)[/tex]
3. Speed of light [tex]\(c = 3.0 \times 10^8 \, \text{m/s}\)[/tex]

We need to convert the mass defect from atomic mass units (amu) to kilograms (kg). After converting, we can then use the mass-energy equivalence formula.

### Step-by-Step Solution:

1. Convert the mass defect to kilograms:
[tex]\[ \Delta m_{kg} = 0.09564 \, \text{amu} \times 1.6606 \times 10^{-27} \, \frac{\text{kg}}{\text{amu}} \][/tex]

2. Apply Einstein’s mass-energy equivalence principle:
[tex]\[ E = \Delta m \cdot c^2 \][/tex]
Using [tex]\(\Delta m_{kg}\)[/tex] from the first step:
[tex]\[ E = (0.09564 \times 1.6606 \times 10^{-27} \, \text{kg}) \times (3.0 \times 10^8 \, \text{m/s})^2 \][/tex]

So the correct setup to calculate the nuclear binding energy is:
[tex]\[ 0.09564 \, \text{amu} \times \left(1.6606 \times 10^{-27} \, \text{kg/amu}\right) \times \left(3.0 \times 10^8\right)^2 \][/tex]

After calculating this expressing (as per above information), the binding energy comes out to be:
[tex]\[ E = 1.429378056 \times 10^{-11} \, \text{J} \][/tex]

Therefore, the correct answer for the setup is:
[tex]\[ 0.09564 \, amu \times \left(1.6606 \times 10^{-27} \, kg \right) / amu \times \left(3.0 \times 10^8\right)^2 \][/tex]