A fusion reactor that uses only deuterium for fuel would have the following two

Question

A fusion reactor that uses only deuterium for fuel would have the following two reactions taking place in the reactor.

The 3H produced in the second reaction would react immediately with another 2H as follows.

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A fusion reactor that uses only deuterium for fuel would have the following two reactions taking place in the reactor. 2H + 2H rightarrow 3He + n + 3.27 MeV 2H + 2H rightarrow 3H + 1H + 4.03 MeV The 3H produced in the second reaction would react immediately with another 2H as follows. 3H + 2H rightarrow 4He + n + 17.6 MeV. The ratio of 2H to 1H atoms in naturally occurring hydrogen is 1.5 times 10-4. How much energy would be produced from 2 L of water if all of the 2H nuclei undergo fusion?

Explanation / Answer

energy released by 5 2H nuciei = (7.3+ 17.6) =24.9 Mev

first we need to calculate Nh than Nd

for calculating nh we ll use the following steps

no. of H atom in 4 L h20 Nh = 2*4/18 * Na= 2.76 *10^26

no. of D atom in 4 L h20 Nd = Nh * 1.5 * 10^-4

= 4.01 * 10^22

hence energy = Nd/5 * Q = 19.97 * 10^22 Mev

= 3.2 *10^10 J

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