Question

1. In this question, you will explore the concept of nuclear fission and how it works to power the Sun. Though there are steps in between, the process begins with 4 protons and ends with a 4 He nucleus and light: 4p → 4He + y

a) The mass of one proton is 1.6726×10-27 kg, while the mass of a 4 He nucleus is 6.643×10-27 kg. Calculate the difference in mass between the two.

b) Using Einstein’s famous equation, E = mc2 , calculate the energy released by this change in mass. Express your answer in Joules (J).

c) Taking the rate of fusion reactions in the Sun to be 8.9×1037 reactions per second, and using your answer to (b), calculate the luminosity (power; energy per time) produced by fusion. Compare your answer to the measured solar luminosity of 3.8×1026 W.

Answer 1

a )

mass difference = product mass - reactant mass

                           = 6.643×10-27 - 1.6726×10-27

                           = 4.9704 x 10^-27 kg

b)

E = m c2

E = 4.9704 x 10^-27 x (3 x10^8)^2

E = 4.473 x 10^-10 J

c )

luminosity = 4.473 x 10^-10 x 8.9 x 10^37 J / s = 3.98 x 10^28 J/s (W)

3.98 x 10^28 W is more than solar luminosity 3.8×10^26 W.