If 5 mg of {}^{235}U is completely destroyed | vedclass

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(A) The number of uranium atoms $N$ in $5 \ mg$ of ${}^{235}U$ is given by $N = \frac{m}{M} 	imes N_A$, where $m = 5 	imes 10^{-3} \ g$, $M = 235 \

Vedclass · Accepted Answer

$4 	imes 10^8 \ J$

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(A) The number of uranium atoms $N$ in $5 \ mg$ of ${}^{235}U$ is given by $N = \frac{m}{M} 	imes N_A$, where $m = 5 	imes 10^{-3} \ g$, $M = 235 \ g/mol$, and $N_A = 6.022 	imes 10^{23} \ atoms/mol$.$N = \frac{5 	imes 10^{-3}}{235} 	imes 6.022 	imes 10^{23} \approx 1.28 	imes 10^{19} \ atoms$.The total energy released $E$ is $N 	imes E_{fission}$, where $E_{fission} = 200 \ MeV = 200 	imes 10^6 	imes 1.6 	imes 10^{-19} \ J = 3.2 	imes 10^{-11} \ J$.$E = 1.28 	imes 10^{19} 	imes 3.2 	imes 10^{-11} \ J \approx 4.096 	imes 10^8 \ J$.Thus, the approximate energy released is $4 	imes 10^8 \ J$.

If $5 \ mg$ of ${}^{235}U$ is completely destroyed in an atom bomb, then the approximate total energy released is (given that the energy released per fission is $200 \ MeV$).

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