
The fission type of warhead of some guided missiles is estimated to be equivalent to 30000 tons of TNT. If $3.5 \times {10^8}\,J$ of energy is released by one ton of exploding TNT, how many fissions occur in the explosion of a warhead? An energy of $200\,MeV$ is released by fission of one atom of ${}^{235}U$.
Answer
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Hint: In order to answer this question, to calculate the number of fissions occurring in the explosion of a warhead? An energy of $200MeV$ is released by fission of one atom of ${}^{235}U$ , we will first find the energy that is released by 30000 tons of TNT.And then we will find the energy released per fission. And then we can easily find the number of fissions occurring.
Complete step by step answer:
Given that, energy released by $1\,ton\,of\,TNT = 3.5 \times {10^8}J$.
So, Energy released by $30000\,tons\,of\,TNT = 30000 \times 3.5 \times {10^8}\,J$.
Energy released per fission $ = 200MeV = 200 \times 1.6 \times {10^{ - 13}}\,J$.
The odd-even impact on the charge distribution of fission fragments is a propensity towards producing fragments with even proton numbers. On the other hand, there is no odd-even influence on fragment mass number distribution. Nucleon pair breakage is to blame for this finding.
$\text{Number of fissions} = \dfrac{\text{Total Energy released}}{\text{Energy released per fission}} \\
\Rightarrow \text{Number of fissions}= \dfrac{{6.5 \times {{10}^{25}}}}{{200}} \\
\therefore \text{Number of fissions}= 3.28 \times {10^{23}} \\ $
Hence, the required number of fissions occurring is $3.28 \times {10^{23}}$.
Note:Fission can happen when a heavy atom's nucleus grabs a neutron or it can happen on its own. Fission of heavy materials is an exothermic reaction that can release significant amounts of usable energy in the form of gamma rays and fragment kinetic energy (heating the bulk material where fission takes place).
Complete step by step answer:
Given that, energy released by $1\,ton\,of\,TNT = 3.5 \times {10^8}J$.
So, Energy released by $30000\,tons\,of\,TNT = 30000 \times 3.5 \times {10^8}\,J$.
Energy released per fission $ = 200MeV = 200 \times 1.6 \times {10^{ - 13}}\,J$.
The odd-even impact on the charge distribution of fission fragments is a propensity towards producing fragments with even proton numbers. On the other hand, there is no odd-even influence on fragment mass number distribution. Nucleon pair breakage is to blame for this finding.
$\text{Number of fissions} = \dfrac{\text{Total Energy released}}{\text{Energy released per fission}} \\
\Rightarrow \text{Number of fissions}= \dfrac{{6.5 \times {{10}^{25}}}}{{200}} \\
\therefore \text{Number of fissions}= 3.28 \times {10^{23}} \\ $
Hence, the required number of fissions occurring is $3.28 \times {10^{23}}$.
Note:Fission can happen when a heavy atom's nucleus grabs a neutron or it can happen on its own. Fission of heavy materials is an exothermic reaction that can release significant amounts of usable energy in the form of gamma rays and fragment kinetic energy (heating the bulk material where fission takes place).
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