A laser-fusion fuel pellet has mass 5.0 mg and consists of equal parts (by mass) of deuterium and tritium. (a) If half the deuterons and an equal number of tritons participate in D-T fusion, how much energy is released? (b) At what rate must pellets be fused in a power plant with 3000 -MW thermal power output? (c) What mass of fuel would be needed to run the plant for 1 year? Compare your answer with the \(3.6 \times 10^{6}\) tons of coal needed to fuel a comparable coal-burning power plant.

Short Answer

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a) The energy released from fusion is \(5.25 \times 10^{10}\) J per pellet. b) A pellet must be fused every 0.0175 s. c) The mass of fuel required to run the plant for 1 year is 9 tons, which is significantly less than the amount of coal required for a similar power output.

Step by step solution

01

Calculate the Energy Released

In the D-T fusion, one deuterium nucleus (D) and one tritium nucleus (T) combine to form one helium nucleus and one neutron. The energy released in this reaction is 17.6 MeV. If half the deuterons and an equal number of tritons participate in D-T fusion, this means \(0.5 \times 10^{-5}\) kg of D and the same amount of T are involved. Since one atom of D or T has a mass of about \(2.67 \times 10^{-27}\) kg, this gives \(0.5 \times 10^{-5} / 2.67 \times 10^{-27} = 1.87 \times 10^{22}\) atoms participating. Therefore, the total energy released is \(1.87 \times 10^{22} \times 17.6 \times 10^{6}\) eV = \(3.28 \times 10^{29}\) eV. Using the conversion \(1 eV = 1.602 \times 10^{-19}\) J, this gives an energy of \(5.25 \times 10^{10}\) J.
02

Determine the Rate of Fusion

A 3000-MW power plant generates \(3000 \times 10^{6}\) W of power. Since power equals energy over time, this means the power plant requires \(5.25 \times 10^{10}\) J every \(5.25 \times 10^{10} / 3000 \times 10^{6} = 1.75 \times 10^{-2}\) s. So a pellet must be fused every 0.0175 s.
03

Calculate the Mass of Fuel

To run the plant for 1 year, the number of pellets needed is \(3.15 \times 10^{7}\) s \(1 year = 3.15 \times 10^{7} s\) divided by 0.0175 s/pellet, which gives \(1.8 \times 10^{9}\) pellets. Thus, the total mass of fuel required is \(1.8 \times 10^{9} \times 5.0 \times 10^{-6}\) kg = 9000 kg = 9 tons. This is significantly less than the \(3.6 \times 10^{6}\) tons of coal required to fuel a comparable coal-burning power plant.

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