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How the Sun Works

If you wanted to go to a store and buy a light bulb as powerful as the Sun, you would have to find one that is 4 x 1026 watts. That's a lot of electricity!

But we know that electricity doesn't power the Sun. The only power source that could give off this much energy for billions and billions of years is nuclear fusion. In this activity, you are going to find out how many fusion reactions a second must occur to keep the Sun burning as bright as it does, and therefore how much Hydrogen fuel it takes to keep the Sun going. This is one gas tank you don't want to have to fill!

    1. Hydrogen fusion in the Sun occurs in three steps called the proton-proton chain. You start with four Hydrogen atoms and you end up with one Helium atom. We can find the quantity of energy released in one of the fusion reactions by comparing the mass of what goes in to the mass of what comes out. Where did the mass go? Into energy! Using Einstein's mass-energy relationship to find out how much energy comes out of a single fusion reaction. For data on Hydrogen, Helium, and anything else you might need use your periodic table and your reference tables. 1 u = 1.6605 x 10-27 kg

Amount of energy released in one fusion reaction = ___________________

2. If the Sun gives off 4 x 1026 watts of energy, how many fusion reactions happen every second?

3. Every reaction eats up 4 Hydrogen atoms. So how much mass does the Sun fuse every second?

4. Finally, if the Sun were 2 x 1030 kg, how long would it take for the Sun to use all of its available mass in the fusion process?

Answer key

Question 1:

Find the mass defect in kilograms and then turn that into energy using E = mc2:

4 Hydrogen atoms - 1 Helium atom = mass defect (kg)
4 (1.673 x 10-27 kg.) - 1 (6.645 x 10-27 kg.) = 0.048 x 10-27 kg.

E = (0.048 x 10-27 kg) (3 x 108 m/s)2 = 4.3 x 10-12Joules
Amount of energy released in one fusion reaction = 4.3 x 10-12Joules/reac.

Question 2:

If the Sun gives off 4 x 1026 watts of energy, how many fusion reactions happen every second?

4 x 1026 Joules/second / 4.3 x 10-12 Joules/react = 9.3 x 1037 reactions/second.

Question 3:

Every reaction fuses 4 Hydrogen atoms:

(9.3 x 1037 reactions/second) x (6.693 x 10-27 kg/reaction) = 6.224 x 1011 kg/s

Question 4:

2 x 1030 kg / 6.224 x 1011 kg/s 3.213 x 1018 seconds = 1011 years!

In reality, the entire Sun is not Hydrogen and fusion only occurs in the core. So about 70% of the Sun is available for fusion, making this number smaller.

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