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).

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.