Conservation of Energy fusion reaction

[crybllrd]

Homework Statement

Consider the following fusion reaction
²H+³H →⁴He + n
in which deuterium and tritium fuse together to form a stable isotope of helium plus a neutron.
In all of physics, we see experimental confirmation of Conservation of Energy, in which the total
amount of energy in a closed system is constant. The total rest plus kinetic energy before the
reaction must be equal to the total rest plus kinetic energy after the reaction.

(a) Look up the masses of the three nuclei and of the neutron in kilograms, keeping at least six
significant figures. Cite your reference. Because we have the same number of electrons before
and after the reaction (and because atomic binding energies are tiny compared to the energies
in this problem), it is not important whether you use nuclear masses or atomic masses.

(b) Convert each of those masses into rest energies in electron-volts (eV). Recall that the speed of
light is 299,792,458 m/s, and that 1 eV = 1.60217653× 10−19 J.

(c) Assuming that the deuterium and tritium have a total of 0.05 MeV of kinetic energy before
the reaction, use conservation of energy to calculate the total kinetic energy after the reaction.
Recall that we handle addition and subtraction differently than multiplication and division
when dealing with significant figures.

(d) Assuming that all of this kinetic energy can be converted with 100% efficiency into usable
energy, how many fusion reactions per second do we need to generate 1 MW of power?

Homework Equations

(a) Nuclei Masses:
Deuterium: 3.32108e-27 kg
Tritium: 4.98162e-27 kg
Isotope of Helium: 6.64215e-27 kg
Neutron: 1.67493e-27 kg

The Attempt at a Solution

(b)First I will convert the masses to eV (online conversion):
Deuterium: 1,862,990,000 eV
Tritium: 2,794,480,000 eV
Isotope of Helium: 3,725,970,000 eV
Neutron: 939,567,000 eV

(c) I will add together the eV of ⁴He + neutron = 4,665,537,000 eV
Now I will convert eV to MeV:
4,665.537 MeV
Can I subtract this by .05 MeV? I feel like I am on the wrong track

 

[dannoshepard]

do you go to msu?

 

[tomcue]

here.

Your calculations for (b) are correct. For (c), you are on the right track but you need to use the masses of the nuclei to calculate the total rest energy before and after the reaction. The total rest energy before the reaction is:
(2 x 1,862,990,000 eV) + (3 x 2,794,480,000 eV) = 14,724,940,000 eV
The total rest energy after the reaction is:
(1 x 3,725,970,000 eV) + (1 x 939,567,000 eV) = 4,665,537,000 eV
Therefore, the total kinetic energy after the reaction is:
14,724,940,000 eV - 4,665,537,000 eV = 10,059,403,000 eV
Converting this to MeV, we get:
10,059,403,000 eV = 10,059 MeV
Therefore, the total kinetic energy after the reaction is 10,059 MeV.

For (d), we need to calculate the number of fusion reactions per second needed to generate 1 MW of power. We can use the formula:
Power = (number of reactions per second) x (energy per reaction)
1 MW = (number of reactions per second) x 10,059 MeV
Rearranging this equation, we get:
(number of reactions per second) = 1 MW / 10,059 MeV
Now we need to convert 1 MW to eV:
1 MW = 1,000,000,000 eV
Therefore:
(number of reactions per second) = 1,000,000,000 eV / 10,059 MeV = 99,409 reactions per second
Therefore, we would need 99,409 fusion reactions per second to generate 1 MW of power.