One atom of Deuterium weighs 2.013553AMU. One atom of Helium weighs 4.002602AMU. In nuclear fusion, the process which powers the Sun and all other stars, 2 atoms of Deuterium are combined to form 1 atom of helium. Calculate the energy released by one such combination. (1AMU = 1.660539x10-27kg)
I really don't know what to do for this question. Even a formula would help me. thanks
E = m c^2
The formula you need is Einstein's famous
E = m c^2
You must have heard of that equation before.
In this case, the m is the mass lost in the reaction
2D -> He
You must use units of kg for the mass loss, not AMU. Then multiply the square the speed of light (in m/s) and you will get the answer in Joules.
Well, I gave you the necessary equation but perhaps I should elaborate.
You were given very precise values for the masses of the two deuterium atoms and the one He atom.
Add the masses of two De atoms and subtract He atom.
That is mass that disappeared, violating the laws of classical physics (conservation of mass) and requiring the use of Einstein's equation relating mass and energy.
To calculate the energy released by the fusion of two atoms of Deuterium to form one atom of Helium, we can use Einstein's famous equation: E = mc^2, where E represents the energy released, m represents the change in mass, and c represents the speed of light.
First, we need to calculate the change in mass. We know the mass of two atoms of Deuterium (2 × 2.013553 AMU) and the mass of one atom of Helium (4.002602 AMU). We need to convert these atomic mass units (AMU) to kilograms using the conversion factor provided: 1 AMU = 1.660539 × 10^-27 kg.
Mass of two atoms of Deuterium = 2 × 2.013553 AMU = 4.027106 AMU
Mass of one atom of Helium = 4.002602 AMU
Converting to kilograms:
Mass of two atoms of Deuterium = 4.027106 AMU × (1.660539 × 10^-27 kg/1 AMU) = 6.692228 × 10^-27 kg
Mass of one atom of Helium = 4.002602 AMU × (1.660539 × 10^-27 kg/1 AMU) = 6.644819 × 10^-27 kg
Now, we can calculate the change in mass:
Change in mass (Δm) = initial mass - final mass
Δm = (6.692228 × 10^-27 kg) - (6.644819 × 10^-27 kg) = 4.739640 × 10^-29 kg
Finally, we can use Einstein's equation to find the energy released:
E = Δmc^2
E = (4.739640 × 10^-29 kg) × (3 × 10^8 m/s)^2
E = 1.342498 × 10^-12 Joules
Therefore, the energy released by the fusion of two atoms of Deuterium to form one atom of Helium is approximately 1.342498 × 10^-12 Joules.
Note: The speed of light, c, is approximately 3 × 10^8 m/s.