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Question: Number of Uranium-235 nuclei required to undergo fission to give \(9 \times {10^{12}}\,J\) of energy...

Number of Uranium-235 nuclei required to undergo fission to give 9×1012 J9 \times {10^{12}}\,J of energy is
A. 2.8125×10242.8125 \times {10^{24}}
B. 28.125×102428.125 \times {10^{24}}
C. 281.25×1024281.25 \times {10^{24}}
D. 28125×102428125 \times {10^{24}}

Explanation

Solution

We know that the energy released in the fission of one uranium nucleus is 200 MeV200\,MeV. If we take the number of nuclei required to produce energy of 9×1012 J9 \times {10^{12}}\,J as nn . Then n times the energy released in the fission of one uranium nucleus will be equal to 9×1012 J9 \times {10^{12}}\,J.
By using this we can find the value of n by dividing the energy 9×1012 J9 \times {10^{12}}\,J with the value of energy produced in the fission of one uranium nucleus.

Complete step by step solution:
We know that the energy produced by fission of one uranium nucleus is
E=200 MeVE = 200\,MeV
We need to find the number of uranium nuclei required to undergo fission to give 9×1012 J9 \times {10^{12}}\,J of energy.
Let us find the energy released in the fission of one uranium nucleus in Joule.
We know that,
1 MeV=106 eV1\,MeV = {10^{6\,}}eV
⇒E=200 MeV=200×106 eV\Rightarrow E = 200\,MeV = 200 \times {10^{6\,}}eV
In order to convert energy in electron volt to joule we need to multiply it by 1.6×10−191.6 \times {10^{ - 19}}.
∵1 eV=1.6×10−19J\because 1\,eV = 1.6 \times {10^{ - 19}}J
Then we get
⇒E=200×106 eV=200×106×1.6×10−19 J\Rightarrow E = 200 \times {10^{6\,}}eV = 200 \times {10^6} \times 1.6 \times {10^{ - 19}}\,J
⇒E=200×1.6×10−13J\Rightarrow E = 200 \times 1.6 \times {10^{ - 13}}J
This is the energy per fission of uranium 235 nuclei.
Let us suppose that nn molecules of uranium undergo fission to give 9×1012 J9 \times {10^{12}}\,J of energy.
That is n times the energy produced per fission is equal to 9×1012 J9 \times {10^{12}}\,J .
⇒n×E=9×1012 J\Rightarrow n \times E = 9 \times {10^{12}}\,J
From this we can calculate the value of n as
⇒n=9×1012E \Rightarrow n = \dfrac{{9 \times {{10}^{12}}}}{E}\,
⇒n=9×1012 J200×1.6×10−13J \Rightarrow n = \dfrac{{9 \times {{10}^{12}}\,J}}{{200 \times 1.6 \times {{10}^{ - 13}}J}}\,
∴n=2.8125×1024 \therefore n = 2.8125 \times {10^{24}}\,
This is a number of uranium nuclei required to undergo fission to give 9×1012 J9 \times {10^{12}}\,J joules of energy.

So, the correct answer is option A.

Note: Fission is the process by which a heavy nucleus splits into smaller lighter nuclei. During the fission less energy is required than the energy needed to bind them together so energy will be released in fission reaction. The energy released in the case of one uranium nucleus is 200 MeV200\,MeV.