In the deuterium-tritium fusion reaction find the rate at which deuterium and tritium are consumed to produce 1 MW. The Q-value of deuterium-tritium reaction is 17.6 MeV. You can assume that the efficiency is 100%.

Solution

Energy released per fusion = 17.6 MeV

Number of fusion reactions to produce 1 MW,

= \frac{10^{6}}{17.6 \times 1.6 \times 10^{- 19} \times 10^{6}} = 3.55 \times 10^{17}

In each reaction one atom of deuterium and one atom of tritium are consumed.

Mass of 3.55 × 10¹⁷ atoms of deuterium consumed per second,

= \frac{2 \times 3.55 \times 10^{17}}{6.023 \times 10^{23}} g = 1.1788 \times 10^{- 6} g / s = 1.179 \times 10^{- 19} kg / s

Mass of tritium consumed per second =

\frac{3 \times 3.55 \times 10^{7}}{6.023 \times 10^{23}} g / s = 1.768 \times 10^{- 6} g / s

= 1.768 \times 10^{- 9} kg / s

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