Answered

When hydrogen nuclei are fused together to form a helium nucleus, the mass of the product is $0.7\%$ less than the mass of the reactants. If 1 gram $\left(1 \times 10^{-3} kg\right)$ of protons were completely fused into helium nuclei, the mass of the helium nuclei produced would be $6.8 \times 10^{-6} kg$ less than the mass of the protons. Using the equation $E=m c^2$, calculate how many joules of energy would be produced by converting $6.8 \times 10^{-6} kg$ of matter into energy in this fusion reaction. Use $c = 3 \times 10^8 m / s$.


A$E = 6.05 \times 10^{11} kg \times m^2 / s^2$
B$E = 6.10 \times 10^{11} kg \times m^2 / s^2$
C$E = 6.12 \times 10^{11} kg \times m^2 / s^2$
D$E = 6.25 \times 10^{11} kg \times m^2 / s^2$
E$E = 6.30 \times 10^{11} kg \times m^2 / s^2$

Verified Answer :

  • Identify the mass defect \(m = 6.8 \times 10^{-6} \text{ kg}\) and the speed of light \(c = 3 \times 10^8 \text{ m/s}\).
  • Apply the formula \(E = mc^2\) for energy calculation.
  • Substitute the values to find \(E = (6.8 \times 10^{-6}) \times (3 \times 10^8)^2 = 6.12 \times 10^{11} \text{ J}\).
  • Conclude that the energy produced is \(\boxed{6.12 \times 10^{11} \text{ kg} \cdot \text{m}^2/\text{s}^2}\).

Explanation

  1. Problem Analysis and Formula Selection

The problem asks for the energy released during a fusion reaction where a mass defect occurs. The mass defect is given as \(m = 6.8 \times 10^{-6} \text{ kg}\), and the speed of light is \(c = 3 \times 10^8 \text{ m/s}\). According to Albert Einstein's mass-energy equivalence principle, the energy \(E\) produced by a change in mass \(m\) is given by the formula:
\[E = mc^2\]
Here, \(E\) is measured in Joules (J), which are equivalent to \(\text{kg} \cdot \text{m}^2/\text{s}^2\). Our task is to calculate this value and select the correct option.

  1. Calculating the Energy Released

We substitute the given values into the equation:
\[E = (6.8 \times 10^{-6} \text{ kg}) \times (3 \times 10^8 \text{ m/s})^2\]
First, we calculate the square of the speed of light:
\[c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \text{ m}^2/\text{s}^2\]
Next, we multiply this by the mass defect \(m\):
\[E = 6.8 \times 10^{-6} \times 9 \times 10^{16}\]
Performing the multiplication:
\[6.8 \times 9 = 61.2\]
And the powers of ten:
\[10^{-6} \times 10^{16} = 10^{10}\]
Combining these results gives:
\[E = 61.2 \times 10^{10} = 6.12 \times 10^{11} \text{ J}\]
Thus, the energy produced is \(6.12 \times 10^{11} \text{ J}\).

  1. Conclusion and Final Answer

Comparing our calculated result \(E = 6.12 \times 10^{11} \text{ kg} \cdot \text{m}^2/\text{s}^2\) with the provided options, we see that it matches the third option perfectly.
The final answer is \(\boxed{6.12 \times 10^{11} \text{ kg} \cdot \text{m}^2/\text{s}^2}\).

Examples

In nuclear physics, understanding the mass-energy equivalence is crucial for estimating the power output of stars and nuclear reactors. For example, knowing how much mass is converted into energy allows scientists to predict the lifespan of the Sun or calculate the fuel requirements for a nuclear power plant, ensuring we can harness clean energy efficiently. This principle is a cornerstone of modern science, bridging the gap between matter and energy.

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