The total power output of the Sun is about 3.8 × 1026 W, and the power-producing core has a radius of about 1.4 × 105 km (about 20% of the full radius of the Sun).
Using these figures and your knowledge of the energy release in the proton– proton chain estimate the power density (i.e. the number of watts per cubic metre) in the core of the Sun, and the number of helium nuclei formed per second per cubic metre. Give your answer in scientific notation and to two significant figures
The second step in the proton–proton chain is the reaction:
2 1 3
H+ H → He
1 1 2
State which nucleons are present in the
3
He nucleus and use the fact that it is
2
stable to estimate the strength of the strong nuclear force between nucleons
In stars greater than about nine times that of the Sun, the production of heavy elements and release of energy by nuclear fusion occurs until iron is formed. Explain why fusion of elements heavier than iron cannot power the star beyond this point
(1) From the energy output of the sun and the volume of the power-producing core, determine the power released per volume in that core.
(2) From the energy release per He-atom creation in the proton-proton chain, and using the answer from (1), compute the number of He atoms created per unit volume per unit time.
(3) Fusion of elements heavier that iron is not possible in stars because the "curve of binding energy" has a minimum there. Google that phrase for more details, if needed. It can happen endothermically in supernova explosions. That is where it is believed all elements heavier than iron were created.
To calculate the power density in the core of the Sun, we need to divide the total power output by the volume of the core.
First, let's convert the radius of the core from kilometers (km) to meters (m):
Radius of core = 1.4 × 10^5 km = 1.4 × 10^8 m
Next, we calculate the volume of the core by using the formula for the volume of a sphere:
Volume of core = (4/3)π(radius of core)^3
Plugging in the values:
Volume of core = (4/3)π(1.4 × 10^8)^3 = 11.5 × 10^24 m^3
Now, we can calculate the power density:
Power density = Total power output / Volume of core
Total power output of the Sun = 3.8 × 10^26 W
Power density = (3.8 × 10^26 W) / (11.5 × 10^24 m^3) = 33.04 × 10^1 W/m^3
Rounded to two significant figures in scientific notation, the power density in the core of the Sun is approximately 3.3 × 10^2 W/m^3.
To determine the number of helium nuclei formed per second per cubic meter, we need to know the rate at which the proton-proton chain reaction occurs. Without that information, we cannot provide a specific numerical answer.
Regarding the second part of your question, the helium-3 nucleus (3He), formed in the proton-proton chain reaction, consists of two protons and one neutron. It is stable due to the strong nuclear force, which is responsible for binding nucleons (protons and neutrons) together in the nucleus of an atom.
Fusion reactions for elements heavier than iron require an input of energy instead of releasing energy. This is because nickel-56 and iron-56 have the highest binding energies per nucleon, meaning they are the most tightly bound. Any fusion process beyond iron requires additional energy to break these highly stable nuclei apart. Therefore, fusion of elements heavier than iron cannot release energy and sustain the star's power beyond a certain point.