Question

In a reactor $U^{235}$ is used. $2\;kg$ is consumed in $30\;days$, and energy fission is $185\;MeV$. Find the power of reactors.

Answer 1

Hint: Nuclear energy or nuclear power is the conversion of heat produced due to nuclear power to produce electricity. Here, the heat produced by the nuclear energy is used to run turbines which in turn produce electricity. The nuclear energy is produced either due to nuclear fission, fusion or nuclear decay.

Formula used:
$\dfrac{E}{T}=P$

Complete step by step answer:
We know that uranium being a heavy and stable element undergoes fission or breaking to give smaller and stable elements. During this process it releases a huge amount of energy which is called nuclear energy which is used to produce electricity. Also during uranium fission the reaction emits gamma rays. In the given nuclear reaction, uranium is the nuclear fuel or the fissile which can produce energy. Also there are two naturally occurring isotopes of uranium, namely $U^{235}$ and $U^{238}$, however among the two, $U^{238}$ is stable.
Given that, $235\;u$ of $U$ produces $185\;MeV$of energy.
Then the energy produced by $1\;u$ is given as $\dfrac{185}{235}=0.787Mev$
We know that $1MeV=1.602\times 10^{-13} J$
Then the energy produced by $1\;u$ is $0.787\times 1.602\times 10^{-13} =1.26\times 10^{-13} J$
Also $1u=1.66\times 10^{-27}kg$
Then, we can say that $1.66\times10^{-27}kg$ produces $1.26\times 10^{-13}J$
Then the energy produced by $2\;kg$ of $U$ is given as $\dfrac{1.26\times 10^{-13}\times 2}{1.66\times 10^{-27}}=1.517\times 10^{14}J$
Then we know that power is the energy consumed per unit time in seconds, here time is given as $30days=30\times24\times 60\times 60=2.59\times 10^{6}s$
Then the power of the reactor is given as $P=\dfrac{E}{t}=\dfrac{1.517\times 10^{14}}{2.59\times 10^{6}}=5.85\times 10^{7}W$

Hence the power consumed by $2\;kg$$U$ is in $30\;days$is $5.85\times 10^{7}W$

Note:
This question has some major conversions, like converting the mass from atomic mass units to kg , converting the energy from MeV to joules and converting time in terms of days to seconds. It is suggested to remember these conversions for easy calculations. Otherwise the sum is quite easy and straightforward.