Use the exact values you enter in previous answer(s) to make later calculation(s

Question

Use the exact values you enter in previous answer(s) to make later calculation(s). The magnitude of the energy produced by the fusion of two deuterium nuclei (^2_1 H is 4.033 MeV. Convert this energy to joules. 6.45 10**-13 In seawater, deuterium has a concentration of 0.0003% relative to that of normal hydrogen (^1_1 H). If all the deuterium in 5.2 g of seawater undergoes fusion, how much energy would be released? (The concentration of H_2O in seawater is 18.66 g of seawater per mol of H_2O molecules. Give your answer as the magnitude of energy released.) 6.0 *10 **4 J

Explanation / Answer

(a) 1 MeV = 1.60218e-13 J

therefore, 4.033 MeV = 4.033 * 1.60218e-13 = 6.46e-13 J

(b) Since, a pair of deuterium atoms is consumed in one fusion event, we first need to calculate the number of deuterium pairs in 5.2g of deuterium

Also, it has been given that the relative concentration of deuterium is 0.0003% that of normal hydrogen

Therefore, in 18.66g/mol of sea water, the relative concentration of deuterium will be = 0.0003 * 18.66 = 5.598e-3 g/mol

Now, the number of deuterium pairs, N = [Msample/2Md]* NA = [5.2/(2*5.598*10-3)] * (6.023*1023) = 2.797e26

Therefore, the total energy released = 2.797*1026 * 6.45*10-13 = 1.8*1014 J

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