Question

Find the amount of energy released in $\text{kWh}$ when $1\text{g}$ of Uranium atoms ${}_{92}{U}^{235}(235.0439$ $\text{amu})$ undergoes fission by slow neutron $\left(1.0087\text{amu}\right)$ and is split into Krypton ${}_{36}K{r}^{92}\left(91.8973\text{amu}\right)$ and Barium ${}_{56}B{r}^{141}\left(140.9139\text{amu}\right)$ (assuming no energy is lost).
$[1\text{amu}{c}^2=931.5\text{MeV}]$.

• A. $2.28\times {10}^4$
• B. $6.3\times {10}^5$
• C. $3\times {10}^6$
• D. $1.8\times 0^5$

Answer 1

The correct option is A $2.28\times {10}^4$

Fission equation:

${}_{92}^{235}U{+}_0^{1}n{\to }_{56}^{141}Ba{+}_{36}^{92}Kr+3{(}_0^{1}n)$

Mass defect:
$\Delta m=\left[\right\{235.0439+1.0087\}-\{140.9139+91.8973+3\times 1.0087\left\}\right]$

$=0.2153\text{amu}$

$\text{Equivalent energy}\left(E_{eqv}\right)=\Delta m{c}^2=\Delta m\times 931\text{MeV}=0.2153\times 931=200.4\text{MeV}$

Number of atoms/fissions in $1\text{g}$ of atom is,

$N_a=\frac{1}{235}\times 6.023\times {10}^{23}=2.56\times {10}^{21}$

Energy released in fission of $1\text{g}$ of ${}_{92}^{235}U$ is,

$=N_a{E}_{eqv}$

$=2.56\times {10}^{21}\times 200.4\text{MeV}$

$=2.56\times {10}^{21}\times 200.4\times 1.6\times {10}^{-13}\text{J}[\because 1e=1.6\times {10}^{-19}\text{C}]$

$=\frac{2.56\times {10}^{21}\times 200\times 1.6\times {10}^{-13}}{3.6\times {10}^6}\text{kWh}$

$\approx 2.28\times {10}^4\text{kWh}$

Hence, $\left(A\right)$ is the correct answer.