Question

About $180\text{MeV}$ of usable energy is released in the neutron induced fissioning of a ${}_{92}{\text{U}}^{235}$ nucleus. If the reactor using ${}_{92}{\text{U}}^{235}$ as fuel continuously generates $100\text{MW}$ power. The time it will take for $1\text{kg}$ of the uranium to be consumed in days .(integer only)

Answer 1

One ${\text{U}}^{235}$ nuclei produces $180\text{MeV}$

Power generated, $P=\frac{N}{t}\times E$

To generate $100\text{MW},{\text{U}}^{235}$ used per second is,

$\frac{N}{t}=\left(\frac{100\times {10}^5}{185\times {10}^6\times e}\right)\times \frac{235}{N_A}{\text{gm s}}^{-1}$

$\therefore$ The time required for $1\text{kg}$

$t={10}^3\times \frac{185\times {10}^6\times e\times N_A}{100\times {10}^6\times 235}=\frac{1850}{235}e{N}_A$

$=\frac{75.82\times {10}^4}{86400}$

$\therefore t=8.77\text{days}\approx 9\text{days}$