Question

Assuming that about $200\text{MeV}$ of energy is released per fission of ${}_{92}{\text{U}}^{235}$ nuclei, then the mass of ${}_{92}{\text{U}}^{235}$ consumed per day in fission reactor of power $1\text{MW}$ will be approximately

• A. ${10}^{-2}\text{g}$
• B. $1\text{g}$
• C. $100\text{g}$
• D. $10\text{g}$

Answer 1

The correct option is B $1\text{g}$

Given,
Power of fission reactor is, $P={10}^6\text{W}={10}^6{\text{Js}}^{-1}$

Time required $=t=1\text{day}=24\times 36\times {10}^2=8.64\times {10}^4\text{s}$

Energy produced, $U=Pt={10}^6\times 8.64\times {10}^4=8.64\times {10}^{10}\text{J}$

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

$200\text{MeV}=200\times {10}^6\times 1.6\times {10}^{-19}=32\times {10}^{-12}\text{J}$

The number of ${}_{92}{\text{U}}^{235}$ atoms used is,

$=\frac{8.64\times {10}^{10}}{32\times {10}^{-12}}=27\times {10}^{20}$

The mass of ${}_{92}{\text{U}}^{235}$ required is,

$m=\left(\frac{235}{6\times {10}^{23}}\right)(27\times {10}^{20})=1.058\text{g}\simeq 1\text{g}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.