Question

The amount of $U^{235}$ to be fissioned, to operate a $10kW$ nuclear reactor is (Approximately)

• A. $1.2\times {10}^{-5}gm/s$
• B. $1.2\times {10}^{-7}gm/s$
• C. $1.2\times {10}^{-9}gm/s$
• D. $1.2\times {10}^{-11}gm/s$

Answer 1

Answer: (B)

$1.2\times {10}^{-7}gm/s$

$U^{235}\text{releases 200 Mev}$

$\text{Let, required}x\text{g of}{U}^{235}$

$10\times {10}^3=\frac{x}{235}\times 6.023\times {10}^{23}\times 200N$

$\times 6\times {10}^{-19}$

$x=1.2\times {10}^{-7}gm/s$