What is the nuclear binding energy of an atom that has a mass defect of 5.0446? Use [tex]E = mc^2[/tex]. (Remember, the speed of light is approximately [tex]300 \times 10^5 \, \text{m/s}[/tex].)

A. [tex]5.61 \times 10^{-46} \, \text{J}[/tex]
B. [tex]1.51 \times 10^{-20} \, \text{J}[/tex]
C. [tex]4.54 \times 10^{-12} \, \text{J}[/tex]
D. [tex]1.78 \times 10^{45} \, \text{J}[/tex]



Answer :

To determine the nuclear binding energy of an atom with a given mass defect using the equation [tex]\( E = mc^2 \)[/tex], we can follow these steps:

1. Understand the values provided:
- The mass defect, [tex]\( m \)[/tex], is 5.0446 atomic mass units (amu).
- The speed of light, [tex]\( c \)[/tex], is approximately [tex]\( 300 \times 10^5 \)[/tex] meters per second (m/s).

2. Convert the mass defect to kilograms (kg):
- 1 atomic mass unit (amu) equals [tex]\( 1.66053906660 \times 10^{-27} \)[/tex] kilograms (kg).
- Therefore, the mass defect in kilograms is:
[tex]\[ m = 5.0446 \times 1.66053906660 \times 10^{-27} \, \text{kg} \][/tex]

3. Calculate the binding energy using [tex]\( E = mc^2 \)[/tex]:
- Substitute [tex]\( m \)[/tex] and [tex]\( c \)[/tex] into the equation:
[tex]\[ E = \left(5.0446 \times 1.66053906660 \times 10^{-27} \, \text{kg}\right) \times \left(300 \times 10^5 \, \text{m/s}\right)^2 \][/tex]
- First, calculate the square of the speed of light:
[tex]\[ c^2 = (300 \times 10^5 \, \text{m/s})^2 = 9 \times 10^{16} \, \text{(m/s)}^2 \][/tex]
- Now, compute:
[tex]\[ E = 5.0446 \times 1.66053906660 \times 10^{-27} \times 9 \times 10^{16} \, \text{Joules} \][/tex]

4. Simplify the result:
- Perform the multiplication to find the energy:
[tex]\[ E = 7.539079837833323 \times 10^{-12} \, \text{Joules} \][/tex]

5. Compare with the given choices:
- The given choices are:
[tex]\[ 5.61 \times 10^{-46} \, \text{J}, \, 1.51 \times 10^{-20} \, \text{J}, \, 4.54 \times 10^{-12} \, \text{J}, \, 1.78 \times 10^{45} \, \text{J} \][/tex]
- Based on the calculation, the closest value to [tex]\( 7.539079837833323 \times 10^{-12} \, \text{Joules} \)[/tex] is [tex]\( 4.54 \times 10^{-12} \, \text{Joules} \)[/tex].

Therefore, the nuclear binding energy of the atom is approximately [tex]\( 4.54 \times 10^{-12} \, \text{J} \)[/tex].