Question

In a reactor, 2 Kg of \[{}_{{\rm{92}}}{\rm{U}}_{}^{{\rm{235}}}\] fuel used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number, N=\[6.023 \times {10^{26}}\] per kilo mole and 1eV=\[1.6 \times {10^{ - 19}}\,{\rm{J}}\] . The power output of the reactor is close to:
A. 60 MW
B. 54 MW
C. 125 MW
D. 35 MW

Answer 1

Hint: Here, to calculate the power output of a reactor. We have to first calculate the number of moles of Uranium reacted and then we have to calculate the total energy produced.

Formula used:
Power of a reactor=\[\dfrac{{{\rm{Energy}}}}{{{\rm{Time}}}}\]

Complete Step by Step Solution:
Let’s first calculate the moles of Uranium that undergoes reaction in the 30 days.
Given that, 2 kg (2000 g)of \[{}_{{\rm{92}}}{\rm{U}}_{}^{{\rm{235}}}\]fuel used in 30 days and the molar mass if Uranium is 235 g.
Moles of Uranium=\[\dfrac{{{\rm{Mass}}}}{{{\rm{Molar}}\,{\rm{mass}}}} = \dfrac{{2000}}{{235}} = 8.51\]
Now, using the moles of Uranium, we have to calculate the number of atoms that undergoes a reaction.
\[{\rm{Number}}\,{\rm{of}}\,{\rm{moles = }}\dfrac{{{\rm{Number}}\,{\rm{of}}\,{\rm{atoms}}}}{{{\rm{Avogadro}}\,{\rm{Number}}}}\]
Number of atoms or nucleus=Avogadro number \[ \times \] number of moles
Number of nucleus=\[ = 6.023 \times {10^{23}} \times 8.51 = 51.26 \times {10^{23}}\]
So, in the fission reaction of Uranium, \[51.26 \times {10^{26}}\]nucleus present.
Now, we have to calculate the energy due to fission. Given that, per fission, energy released is 200 MeV and 1eV=\[1.6 \times {10^{ - 19}}\,{\rm{J}}\]
Energy\[ = 51.26 \times {10^{23}} \times 200 \times 1.6 \times {10^{ - 19}}\,{\rm{J}}\]
Energy\[ = 102.5 \times {10^{25}} \times 1.6 \times {10^{ - 19}}\,\,{\rm{J}}\]
Energy\[ = 164 \times {10^6}\,{\rm{J}}\]
Now, we have to calculate the time 30 days into second.
Time=\[ = 30 \times 24 \times 60 \times 60\,{\mathop{\rm s}\nolimits} \]
Now, we have to find out the power output.
\[ = \dfrac{{164 \times {{10}^6}\,J}}{{30 \times 24 \times 60 \times 60\,s}}\]
\[ = 6.33 \times {10^{ - 5}}\,J/s\]
\[ = 63.3 \times {10^{ - 6}}\,Watt = 63\,MW\] (1 J/s=1 Watt and 1 watt=\[{10^{ - 6}}\,{\rm{MW}}\] )
Therefore, the power output of the reactor is 63 MG (close to option a). Hence, option A is correct.

Note: It is to be noted that in a fission reaction, the splitting of larger atoms into smaller atoms of two or more than two takes place.