Question

A nuclear reactor generates power at $50$% efficiency by fission of ${}_{92}{U}^{235}$ into equal fragments of ${}_{46}P{d}^{116}$ with the emission of two $\gamma$ -rays of $5.2MeV$ each and three neutrons. The average B.E. per particle of $U^{235}$ and $P{d}^{116}$ is 7.2 MeV and 8.2 MeV respectively. The amount of $U^{235}$ consumed per hour to produce $1600MW$ power is

• A. $127.7gm$
• B. $1.4kg$
• C. $140.5gm$
• D. $281gm$

Answer 1

Answer: (C)

$140.5gm$

Fission process is: ${}_{92}^{235}U\to 2{(}_{46}^{116}Pd)+3{(}_0{1}^{1}n)+2\gamma$

Energy released in the fission process:
$\Delta E=\left[2\right(8.2\times 116)-(7.2\times 235)-2(5.2\left)\right]MeV$
$=200MeV=3.2\times {10}^{-11}J$

Power of reactor, $P=1600\times {10}^{6}J/s$

Number of fissions required to produce this power is:
$N=\frac{P}{\Delta E}=\frac{1600\times {10}^6}{3.2\times {10}^{-11}}=5\times {10}^{19}/s$

Number of Uranium nuclei, $N=5\times {10}^{19}/s$

Mass of Uranium, $m=\left(\frac{235}{6.02\times {10}^{23}}\right)\times 5\times {10}^{19}g=1.95\times {10}^{-2}g/s$

Mass of uranium required per hour $=1.95\times {10}^{-2}\times 3600g/h$

As the efficiency of the reactor is 50% the actual amount of uranium is $70.27\times 2=140.54g$