Here is the complete question:
Calculate the energy released in the following fusion reaction. The masses of the isotopes are: 14N (14.00307 amu), 32S (31.97207 amu), 12C (12.00000 amu), and 6Li (6.01512 amu).
¹⁴N + ¹²C + ⁶Li ⇒ ³²S
Answer:
68.7372 × 10⁻¹⁶ kJ
Explanation:
Given that the reaction; ¹⁴N + ¹²C + ⁶Li ⇒ ³²S
To calculate for the energy released; we need to determine the mass defect (md) of the reaction and which is given as :
Mass defect (md) = [mass of reactants] -[mass of product]
Mass defect (md) = [ ¹⁴N + ¹²C + ⁶Li ] - [ ³²S ]
Mass defect (md) = [ 14.00307 + 12.00000 + 6.01512 ] amu - [ 31.97207 ] amu
Mass defect (md) = 32.01819 - 31.97207
Mass defect (md) = 0.04612 amu
Having gotten the value of our Mass defect (md); = 0.04612 amu
We know that if 1 amu ⇒ 931.5 Mev of energy
∴ 0.04612 amu = 0.04612 × 931.5 Mev of energy
= 42.96078 Mev of energy
where M = million = 10⁶
1 ev = 1.6 × 10⁻¹⁹ Joules
∴ 42.96078 Mev of energy = 42.96078 × 10⁶ × 1.6 × 10⁻¹⁹ J
= 68.7372 × 10⁻¹³ J
= 68.7372 × 10⁻¹⁶ kJ
Hence; the energy released in the above fusion reaction = 68.7372 × 10⁻¹⁶ kJ.
