Question

Assuming $200MeV$ of energy is released per fission of ${}_{92}{U}^{235}$, calculate the energy released when $1kg$ of ${}_{92}{U}^{235}$ undergoes complete fission. What percentage of mass of ${}_{92}{U}^{235}$ gets converted into energy ?

• A. $5.126\times {10}^{26}MeV$; 0.5 %
• B. $5.126\times {10}^{26}MeV$; 0.1 %
• C. $4.126\times {10}^{26}MeV$; 1.0 %
• D. $8.126\times {10}^{26}MeV$; 2.0 %

Answer 1

The correct option is B $5.126\times {10}^{26}MeV$; 0.1 %

$235g$ of ${}_{92}{U}^{235}$ contains Avogadro number $(6.023\times {10}^{23})$ of atoms.
Number of atoms of ${}_{92}{U}^{235}$ in $1kg$ of uranium $=1000\times \frac{6.023\times {10}^{23}}{235}=2.563\times {10}^{24}$
Energy released $=$ Number of atoms $\times$ energy released per fission.
$=2.563\times {10}^{24}\times 200MeV=5.126\times {10}^{26}MeV$
The mass that liberates the above energy can be calculated using $E=m{c}^2$ ......(2)
where $E=5.126\times {10}^{26}\times 1.6\times {10}^{-13}J=8.2\times {10}^{13}Jc=3\times {10}^{8}m{s}^{-1}$
Substituting in (2), we get $m=\frac{E}{c^2}=\frac{8.2\times {10}^{13}}{9\times {10}^{16}}=9.11\times {10}^{-4}kg$
% of mass $=\frac{9.11\times {10}^{-4}}{1}\times 100=0.0911$% $=0.1$ % approximately.