Answer: the required mass is 1.7628 × 10²⁵ μ
Explanation:
Given that;
energy released in the fission of one ²³⁵U₉₂ is 206.6 MeV
power p = 28.7 Mega Watts = 28.7 × 10⁶ W = 28.7 × 10⁶/ 1.6× 10⁻¹³ = 17.9375 × 10¹⁹ MeV/s
now fission need per second will be;
⇒ power / energy released i fission
= 17.9375 × 10¹⁹ MeV / 206.6 MeV = 8.68 × 10¹⁷ per second
now fission need per day will be;
⇒ ( 8.68 × 10¹⁷ × 360 × 24 ) = 7.5 × 10²² per day
hence mass of ²³⁵U₉₂ that is consumed in one day in this reactor will be;
⇒ (235 × 7.5 × 10²²)μ
= 1.7628 × 10²⁵ μ
Therefore the required mass is 1.7628 × 10²⁵ μ