Question

${}_{92}{U}^{235}$ undergoes nuclear fission as follows $n+$ ${}_{92}{U}^{235}\to$ ${}_{42}M{o}^{98}+$ ${}_{54}X{e}^{136}+2n$
Energy released in fission of $1$ gm of ${}_{92}{U}^{235}$ is (masses of n, ${}_{92}{U}^{235}$, ${}_{42}M{o}^{98}$, and ${}_{54}X{e}^{136}$ are $1.0087$, $235.0439,97.9054$ and $135.9170$(all in amu) respectively)

• A. $5.06\times {10}^{23}$ J
• B. $5.06\times {10}^{26}$ J
• C. $8.1\times {10}^7$ J
• D. $8.1\times {10}^{10}$ J

Answer 1

The correct option is D $8.1\times {10}^{10}$ J

Defect of mass during fission of 1 Uranium atom is $U+n-(Mo+Xe+2n)=\Delta m$

or $\Delta M=235.0439+1.0087-(97.9054+135.9170+2\times 1.0087)=0.2128amu=0.2128\times 1.67\times {10}^{-27}kg$

So energy released by $1$ atom is $\Delta m\times (3\times {10}^8{)}^{2}Joule=3.198\times {10}^{-11}Joule$

Number atoms in $\text{1 gm is Avogadro number * 1/a.wt}$= $6.02\times {10}^{23}/235=.0256\times {10}^{23}$

So energy released by $1gm$ will be $3.198\times {10}^{-11}\times .0256\times {10}^{23}=8.18\times {10}^{10}Joule$

Option D is correct.