Determine the power output of a $$_{92}{U^{235}}$$ reactor if it takes 30 days to use $$2kg$$ of fuel. Energy released per fission is $$200\,MeV$$ and $$N = 6.023 \times {10^{26}}$$ per kilomole.
A.
$$63.28\,MW$$
B.
$$3.28\,MW$$
C.
$$0.6\,MW$$
D.
$$50.12\,MW$$
Answer :
$$63.28\,MW$$
Solution :
Number of atoms in $$2kg$$ fuel
$$ = \frac{2}{{235}} \times 6.023 \times {10^{26}} = 5.12 \times {10^{24}}$$
number of fission per second
$$ = \frac{{5.12 \times {{10}^{24}}}}{{30 \times 24 \times 60 \times 60}} = 1.978 \times {10^{18}}$$
Energy released per fission
$$ = 200\,MeV = 200 \times 1.6 \times {10^{ - 13}} = 3.2 \times {10^{ - 11}}J$$
Power output $$ = 3.2 \times {10^{ - 11}} \times 1.978 \times {10^{18}}$$
$$ = 63.28 \times {10^6}W = 63.28\,MW$$
Releted MCQ Question on Modern Physics >> Atoms or Nuclear Fission and Fusion
In the nuclear fusion reaction
$$_1^2H + _1^3H \to _2^4He + n$$
given that the repulsive potential energy between the two
nuclei is $$ \sim 7.7 \times {10^{ - 14}}J,$$ the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s Constant $$k = 1.38 \times {10^{ - 23}}J/K$$ ]
The binding energy per nucleon of deuteron $$\left( {_1^2H} \right)$$ and helium nucleus $$\left( {_2^4He} \right)$$ is $$1.1\,MeV$$ and $$7\,MeV$$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is