Question

The mass of ${}_{92}^{235}$U that has to undergo fission each day to provide 3000 MW of power each day is:

• A. 3.2 g
• B. 320 g
• C. 3.2 kg
• D. 32 kg

Answer 1

The correct option is C 3.2 kg

Each fission released 200 Mev of energy.
Number of fission to take place in a sec to give $3000MW$ power $=\frac{3000\times {10}^6}{200\times 1.6\times {10}^{-19}\times {10}^6}$

$=\frac{3\times {10}^9}{3.2\times {10}^{-11}}=9.375\times {10}^{19}$
Mass of ${}_{92}^{235}U$ required $=1.66\times {10}^{-27}\times 9.375\times {10}^{19}\times 235\times 3600\times 24kgs=3.159kgs$ $\approx 3.2kgs.$