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Consider the following nuclear reaction ...

Consider the following nuclear reaction
`._(1)^(2)D+._(1)^(2)_Dto._(1)^(3)D+._(1)^(1)P`
Given that `m(_(1)^(3)T)=3.01605` amu,
`m(_(1)^(3)p)=1.00728` amu
The mass of deuterium `(._(1)^(2)D)` required per day in order to produce a power output of `10^(9)W` (with `50%` efficiency) is `n.3` kg where `n` is a digit. Find `n`

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Verified by Experts

The correct Answer is:
1
Required mass `=(10^(9))/(Q-"value" xx0.5)xx2xxm(_(1)^(2)D)xx24xx3600=1.3kg`
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