If the Sun converts 5 × 10¹¹ kg of H to He per second,and the mass of a single hydrogen nucleus is 1.7 × 10⁻²⁷ kg,how many net proton-proton reactions go on per second in the Sun? What is the luminosity produced if the mass difference between a single helium nucleus and four hydrogen nuclei is 4 × 10⁻²⁹ kg? Note that 1 watt (W)= 1 m² kg/s³.

Question

Quizplus · Accepted Answer

To find the net proton-proton reactions per second, we need to divide the total mass of hydrogen converted per second by the mass of a single hydrogen nucleus:

(5 × 10¹¹ kg)/(1.7 × 10⁻²⁷ kg) = 2.94 × 10³⁷

This is the total number of proton-proton reactions per second, but we need to divide by two since each reaction involves two protons. Therefore:

Net proton-proton reactions per second = (2.94 × 10³⁷)/2 = 1.47 × 10³⁷

To find the luminosity produced, we need to convert the mass difference into energy using Einstein's famous equation:

E = mc²

where E is energy, m is mass, and c is the speed of light. The mass difference is:

(4 × 10⁻²⁹ kg) - 4(1.7 × 10⁻²⁷ kg) = -2.36 × 10⁻²⁹ kg

The negative sign indicates that energy is released. The energy released per reaction is:

(-2.36 × 10⁻²⁹ kg)c² = -2.12 × 10⁻¹² J

Again, we need to divide by two since each reaction releases this amount of energy. Therefore:

Energy released per proton-proton reaction = (-2.12 × 10⁻¹² J)/2 = -1.06 × 10⁻¹² J

Finally, we can calculate the luminosity as the product of the energy released per reaction and the net proton-proton reactions per second:

Luminosity = (-1.06 × 10⁻¹² J)(1.47 × 10³⁷) = -1.56 × 10²⁵ W

The negative sign can be ignored and the answer expressed in scientific notation given in the answer choices is closest to 3 × 10¹²⁵ W, which matches Choice A. Therefore, the answer is A.

If the Sun Converts 5 × 10¹¹ Kg of H