A reactor converts $$100\% $$ of given mass into energy and that it operates at a power of $$9 \times {10^7}watt.$$ The mass of the fuel consumed in $$30\,\min$$ in the reactor will be :
A.
$$12 \times {10^{ - 3}}kg$$
B.
$$25 \times {10^{ - 8}}kg$$
C.
$$18 \times {10^{ - 7}}kg$$
D.
$$11 \times {10^{ - 4}}kg$$
Answer :
$$18 \times {10^{ - 7}}kg$$
Solution :
the reactor will conume $$\frac{{9 \times {{10}^7}}}{{9 \times {{10}^{16}}}} \times 1800\,kg$$
$$ = 18 \times {10^{ - 7}}kg$$
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