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Assuming that about `20 M eV` of energy is released per fusion reaction `._(1)H^(2)+._(1)H^(3)rarr._(0)n^(1)+._(2)He^(4)`, the mass of `._(1)H^(2)` consumed per day in a future fusion reactor of powder `1 MW` would be approximatelyA. `0.1 gm`B. `0.01 gm`C. `1 gm`D. `10 gm` |
Answer» Correct Answer - A no of moles of `._(1)H^(2)` consumed `=(1MWxx(24xx3600)sec//day)/((20 MeVxx6.023xx10^(23)))=0.05` `:. m=0.1 g` |
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