## Question

A possible fusion reaction is _{1}^{2}H + _{1}^{2}H —-> _{1}^{3}H + _{1}^{1}H + Q, where Q is the energy released as a result of the reaction. If Q = 4.03 eV, Calculate the atomic mass of _{1}^{3}H in atomic mass units. [_{1}^{2}H = 2.0141u, _{1}^{1}H = 1.00783u, 1u = 931MeV]

### Answer

_{1}^{2}H + _{1}^{2}H —-> _{1}^{3}H + _{1}^{1}H + Q

Let X = atomic of _{1}^{3}H in u,

1u = 931MeV

Therefore, 4.03MeV = (4.03u/931) = 0.004329u

Substitute given values into the nuclear equation to obtain:

2.01410 + 2.01410 —-> x + 1.00783 + 0.004329

Hence, 4.0282 = x + 1.01216

**Therefore, x = 4.0282 – 1.01216 = 3.01604u**