Question

Calculate the amount of energy evolved (in kJ) when 2.50 mg of 2H undergoes fusion according to the equation below. 2H + 3H = 4He + 1n ?

Ive calculated the mass defect through guidance of a video. The video calculates the energy for one mole of 4He. How do i calculate the energy for 2H?

Answer 1

The energy change is for the 'overall' reaction, not just deuterium or tritium. For a 2.50 mg sample of Deuterium (H-2) `Delta E ~ -4.22xx10^6j`.

Explanation:

For the reaction `""_1^2H` + `""_1^3H` => `""_2^4He` + `""_0^1n` determine the nuclear masses ...

Nuclear mass of `""_1^2H` = Mass of `""_1^2H` - Mass of `1e^-`
= 2.01400 amu - 0.000549 amu = 2.013451 amu

Nuclear mass of `""_1^3H` = Mass of `""_1^3H` - Mass of `1e^-`
= 3.01605 amu - 0.000549 amu = 3.015501 amu

Nuclear mass of `""_2^4He` = Mass of `""_2^4He` - mass of 2`e^-`
= 4.00260 amu - 2(0.000549) amu = 4.001502 amu

Mass of 1 `""_1^on` = 1.00865 amu

Change in nuclear mass
`Delta m` = `Sigma(Product Masses)` - `Sigma(Reactant Masses)`
= `[4.001502"amu" + 1.008665"amu"]` - `[2.013451"amu" + 3.015501"amu"]` = `-0.018785 "amu"`
= `-1.8785xx10^-5Kg`

Energy Change for one mole `""_1^2H` ...
`DeltaE = mc^2` = `(-1.8785xx10^-5Kg)(3xx10^8"m/s")^2`
= `-1.68839xx10^12 Kgm^2/s^2` = `-1.68839xx10^12"joules"`

Energy Change for 2.50 mg `""_1^2H` = `2.5xx10^-6 "mole"""_1^2H`
= `2.5xx10^-6cancel("mole")` `(-1.68839xx10^12j/cancel("mole"))`
= `-4.22xx10^6j`

Energy Change in MeV per nucleon ...
= `((-4.22xx10^6cancel(j))/(2.5xx10^-6cancel("mole")))` `((1cancel("mole"))/(6.023xx10^23"nuclei"))` `((1"MeV")/(1.602xx10^-13cancel(j)))`
= `-17.5"MeV"/"nuclei"`