Problem 49

  One proposed fusion reaction combines lithium ⁶₃Li (6.015 u) with deuterium ²₁H (2.014 u) to give helium ⁴₂He (4.003 u): $^6_3Li + \,^2_1H \longrightarrow 2^4_2He$.How many kilograms of lithium would be needed to supply the energy needs of one household for a year, estimated to be $3.8 \times 10^{10}$ J?

SOLUTION: First, we find the energy released per reaction. Then we'll find the total amount of lithium needed to produce that much energy. The missing mass in the reaction is

$$\Delta m = 2.014 u + 6.015 u - 2(4.003 u) = 0.023 u .$$

The equivalent of this in energy (converted to Joules) is

$$\Delta E = (21 MeV)\left (\frac{1.4924 \times 10^{-10} J}{931.5 MeV} \right )$$

$$\Delta E = 3.4 \times 10^{-12} J$$

That's the energy per reaction (or energy per lithium nucleus). The number of nuclei in 1 kg of lithium is

$$N = (1000 g)\left (\frac{6.022 \times 10^{23} nuclei/mol} {6.015 g/mol}\right )$$

N = 10²⁶ nuclei

Therefore, 1 kg of lithium would produce an energy of

$$E = (3.4 \times 10^{-12} J/nuclei)(10^{26} nuclei)$$

$$E = 3.4 \times 10^{14} J$$

If the energy needed per household per year is $3.8 \times 10^{10}$ J, then the amount of lithium required would be

$$m = \frac{3.8 \times 10^{10} J}{3.4 \times 10^{14} J/kg}$$

$$m = 1.1 \times 10^{-4} kg$$