Question

The amount of $U^{235}$ in kg which is to be used per hour in a nuclear reactor of capacity $100$ ($E=200MeV/fission$)

• A. $0.43\times {10}^{-1}$
• B. $4.3\times {10}^{-1}$
• C. $4.3\times {10}^1$
• D. $43\times {10}^{-1}$

Answer 1

Answer: (B)

$4.3\times {10}^{-1}$

$\begin{array}{c}\text{given,}\\ \text{capacity of reactor}=100\text{kwatt}\\ =100\text{K Joule/sec}\end{array}$

$\begin{array}{cc}\text{Now, Energy in one hour}& =100\times {10}^3\times 60\times 60\text{Joule}\\ & =36\times {10}^7\text{Joule}\end{array}$

$\begin{array}{c}\text{Let}x\text{kg of}{U}^{235}\text{required to get this energy}\\ \text{the}\text{mole}=\frac{x\times 1000}{235}\end{array}$

$\begin{array}{c}\text{number of atom}=\frac{x\times 1000}{235}\times N_a\text{(avogadro)}\\ \therefore 1\text{atom releases}200\text{Mev energy}\end{array}$

$\frac{x\times 1000\times 6.022\times {10}^{23}\times {10}^6\times 1.6\times {10}^{-19}}{235}=36\times {10}^7$

$\begin{array}{c}x=439.01\times {10}^{-3}kg\\ x=4.3\times {10}^{-1}kg\end{array}$