The Sun emits energy at an average rate of 4.1 x 10^8 J s^-1 - Scottish Highers Physics - Question 8 - 2019

To find the energy released, we first need to calculate the mass lost in the reaction:

1. Calculate the mass before the reaction:

   • Before: Mass of $^{1}H + ^{1}H = 2 \times 3.3436 \times 10^{-27} = 6.6872 \times 10^{-27} \text{ kg}$

2. Calculate the mass after the reaction:

   • After: Mass of $^{2}He + ^{0}n = 5.0082 \times 10^{-27} + 1.6749 \times 10^{-27} = 6.6831 \times 10^{-27} \text{ kg}$

3. Calculate the mass lost:

   • Mass Lost = Mass before - Mass after = $6.6872 \times 10^{-27} - 6.6831 \times 10^{-27} = 4 \times 10^{-10} \text{ kg}$

4. Calculate the energy released using Einstein's mass-energy equivalence:

   $E = mc^2$

   Here, $c = 3.00 \times 10^8 \text{ m/s}$:

   $E = (4 \times 10^{-10}) \times (3.00 \times 10^{8})^2$ $E = 4 \times 10^{-10} \times 9.00 \times 10^{16} = 3.60 \times 10^{7} \text{ J}$

To calculate the number of reactions required:

1. Use the average energy output of the Sun:

   • Average energy output = $4.1 \times 10^{8} \text{ J/s}$

2. Divide the average energy output by the energy released per reaction:

   $\text{Number of reactions} = \frac{\text{Average Energy Output}}{\text{Energy Released per Reaction}}$

   $\text{Number of reactions} = \frac{4.1 \times 10^{8}}{3.60 \times 10^{7}}$

   $\text{Number of reactions} \approx 11.4 \text{ reactions/s}$

   Therefore, approximately 11 reactions would be required per second.