To determine the nuclear binding energy for an atom of helium-4 given the mass defect and using the famous equation [tex]\( E = mc^2 \)[/tex], follow these steps:
1. Identify the given values:
- Mass defect, [tex]\( m = 5.0531 \times 10^{-29} \)[/tex] kilograms
- Speed of light, [tex]\( c = 3 \times 10^8 \)[/tex] meters per second
2. Understand the equation:
The equation [tex]\( E = mc^2 \)[/tex] calculates the energy (E) where:
- [tex]\( m \)[/tex] is the mass defect
- [tex]\( c \)[/tex] is the speed of light
3. Perform the calculation:
Substitute the given values into the equation.
[tex]\[
E = (5.0531 \times 10^{-29} \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2
\][/tex]
Simplify the expression inside the parentheses first:
[tex]\[
(3 \times 10^8 \, \text{m/s})^2 = 9 \times 10^{16} \, \text{ (m}^2/\text{s}^2)
\][/tex]
Now, multiply the mass defect by this squared speed of light:
[tex]\[
E = 5.0531 \times 10^{-29} \, \text{kg} \times 9 \times 10^{16} \, \text{m}^2/\text{s}^2
\][/tex]
4. Combine the numbers:
Multiply the coefficients and add the exponents:
[tex]\[
E = 5.0531 \times 9 \times 10^{-29 + 16} \, \text{J}
\][/tex]
5. Calculate the coefficient:
[tex]\[
5.0531 \times 9 = 45.478
\][/tex]
Combine the exponents:
[tex]\[
10^{-29 + 16} = 10^{-13}
\][/tex]
Therefore:
[tex]\[
E = 45.478 \times 10^{-13} \, \text{J}
\][/tex]
6. Convert to scientific notation:
Move the decimal place one position to the left:
[tex]\[
E = 4.5478 \times 10^{-12} \, \text{J}
\][/tex]
Rounding to two significant figures if necessary, we have approximately:
[tex]\[
E \approx 4.53 \times 10^{-12} \, \text{J}
\][/tex]
Finally, match this result with the options given:
A. [tex]\( 4.53 \times 10^{-15} \)[/tex] joules
B. [tex]\( 4.53 \times 10^{-12} \)[/tex] joules
C. [tex]\( 4.53 \times 10^{12} \)[/tex] joules
D. [tex]\( 4.53 \times 10^{29} \)[/tex] joules
Therefore, the correct answer is:
B. [tex]\( 4.53 \times 10^{-12} \)[/tex] joules
