
What is the power output of $_{92}{Y^{235}}$ reactor if it takes $30$ days to use up $2\,kg$ fuel and if each fission gives $188\,MeV$ of usable energy?
A. $59\,MW$
B. $51 \times {10^4}\,MW$
C. $188\,MW$
D. None of the above
Answer
515.4k+ views
Hint: In order to solve this question we need to understand radioactivity decay. Radioactivity decay is the process in which an unstable atomic nucleus loses energy by radiation. There are three types of rays which emanate from the nucleus, one is alpha ray which is nucleus of helium it is less penetrating in power, next is beta radiation in which electron and positron emanate from nucleus and gamma radiation which is highly penetrating in nature but it has no charge.
Complete step by step answer:
We have to first calculate what is mass of $_{92}{Y^{235}}$ used per reaction is, $m$
Since the total fuel used is , $E = 2kg$ in total time $T = 30\,days$.
So mass used is,
$m = \dfrac{{2 \times {{10}^3}}}{{30 \times 24 \times 60 \times 60}}$
$\Rightarrow m = 7.72 \times {10^{ - 4}}g{\sec ^{ - 1}}$
Since the Avogadro number is, ${N_A} = 6.62 \times {10^{23}}$.
Number of fission reactions per second is, $N = \dfrac{{{N_A}}}{M} \times m$.
Here $M = 235$ is the mass of the reactor.
Putting values we get,
$N = \dfrac{{6.62 \times {{10}^{23}}}}{{235}} \times 7.72 \times {10^{ - 4}}$
$\Rightarrow N = 2.17 \times {10^{18}}{\sec ^{ - 1}}$
So the power of reactor is given by,
$P = N \times e$
Here $e$ is energy that it gives in each reaction, given in question,
$e = 188MeV{\sec ^{ - 1}}$
So putting values we get,
$P = 2.17 \times {10^{18}}{\sec ^{ - 1}} \times 188\,MeV{\sec ^{ - 1}}$
Since $1eV = 1.6 \times {10^{ - 19}}J$ and $1M = {10^6}$
So power of reactor is,
$P = 2.17 \times {10^{18}} \times 188 \times {10^6} \times 1.6 \times {10^{ - 19}}\,W \\$
$\Rightarrow P = 65.27 \times {10^6}\,W \\$
$\therefore P = 65.27\,MW$
So the correct option is D.
Note: It should be remembered that nuclear reaction is of two types one is nuclear fission and fusion. Fission is defined as a nuclear reaction in which two small nuclei combine to form a large nucleus and thereby radiating heat. Nuclear fusion is defined as a nuclear reaction in which a large nucleus breaks into a small nucleus and also some particles like neutrino and antineutrino etc.
Complete step by step answer:
We have to first calculate what is mass of $_{92}{Y^{235}}$ used per reaction is, $m$
Since the total fuel used is , $E = 2kg$ in total time $T = 30\,days$.
So mass used is,
$m = \dfrac{{2 \times {{10}^3}}}{{30 \times 24 \times 60 \times 60}}$
$\Rightarrow m = 7.72 \times {10^{ - 4}}g{\sec ^{ - 1}}$
Since the Avogadro number is, ${N_A} = 6.62 \times {10^{23}}$.
Number of fission reactions per second is, $N = \dfrac{{{N_A}}}{M} \times m$.
Here $M = 235$ is the mass of the reactor.
Putting values we get,
$N = \dfrac{{6.62 \times {{10}^{23}}}}{{235}} \times 7.72 \times {10^{ - 4}}$
$\Rightarrow N = 2.17 \times {10^{18}}{\sec ^{ - 1}}$
So the power of reactor is given by,
$P = N \times e$
Here $e$ is energy that it gives in each reaction, given in question,
$e = 188MeV{\sec ^{ - 1}}$
So putting values we get,
$P = 2.17 \times {10^{18}}{\sec ^{ - 1}} \times 188\,MeV{\sec ^{ - 1}}$
Since $1eV = 1.6 \times {10^{ - 19}}J$ and $1M = {10^6}$
So power of reactor is,
$P = 2.17 \times {10^{18}} \times 188 \times {10^6} \times 1.6 \times {10^{ - 19}}\,W \\$
$\Rightarrow P = 65.27 \times {10^6}\,W \\$
$\therefore P = 65.27\,MW$
So the correct option is D.
Note: It should be remembered that nuclear reaction is of two types one is nuclear fission and fusion. Fission is defined as a nuclear reaction in which two small nuclei combine to form a large nucleus and thereby radiating heat. Nuclear fusion is defined as a nuclear reaction in which a large nucleus breaks into a small nucleus and also some particles like neutrino and antineutrino etc.
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