Question

$1.00\text{kg}$ of ${}^{235}\text{U}$ undergoes fission process. If energy released per event is $200\text{MeV}$, then the total energy released is

• A. $5.12\times {10}^{24}\text{MeV}$
• B. $6.02\times {10}^{23}\text{MeV}$
• C. $5.12\times {10}^{26}\text{MeV}$
• D. $6.02\times {10}^{26}\text{MeV}$

Answer 1

The correct option is C $5.12\times {10}^{26}\text{MeV}$

Given that, the energy released per fission reaction of ${}^{235}\text{U}$,

$E_1=200\text{MeV}$

The number of nuclei in $1\text{kg of}{}^{235}\text{U}$ is given by,

$N=\frac{N_A}{M_o}m$

Where, $M_o\to$ Molecular mass of ${}^{235}\text{U}$
$m\to$given mass.

$\Rightarrow N=\frac{N_A}{235}\times (1\times {10}^3)$

$\Rightarrow N=\frac{6.023\times {10}^{23}}{235}\times {10}^3$

$\Rightarrow N=2.56\times {10}^{24}$

Therefore, the total energy released is given by,

$E=N{E}_1$

$\Rightarrow E=2.56\times {10}^{24}\times 200\text{MeV}$

$\Rightarrow E=5.12\times {10}^{26}\text{MeV}$

Hence, option $\left(C\right)$ is correct.

Why this question? Total energy released = (Number of fission events)$\times$(Energy released in one fission event)