U^{235} nuclear reactor generates energy at a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Total energy required $E = P 	imes t = (3.70 	imes 10^7 	ext{ J/s}) 	imes (144 	imes 10^4 	ext{ s}) = 5.328 	imes 10^{13} 	ext{ J}$.Energy

Vedclass · Accepted Answer

$0.705$

Vedclass · Answer

(B) Total energy required $E = P 	imes t = (3.70 	imes 10^7 	ext{ J/s}) 	imes (144 	imes 10^4 	ext{ s}) = 5.328 	imes 10^{13} 	ext{ J}$.Energy released per fission $E_f = 185 	ext{ MeV} = 185 	imes 10^6 	imes 1.6 	imes 10^{-19} 	ext{ J} = 2.96 	imes 10^{-11} 	ext{ J}$.Number of fissions $N = E / E_f = (5.328 	imes 10^{13}) / (2.96 	imes 10^{-11}) = 1.8 	imes 10^{24} 	ext{ atoms}$.Mass of fuel $m = (N / N_A) 	imes M = (1.8 	imes 10^{24} / 6 	imes 10^{23}) 	imes 235 	ext{ g} = 0.3 	imes 235 	ext{ g} = 70.5 	ext{ g} = 0.0705 	ext{ kg}$.Re-evaluating the calculation: $N = 1.8 	imes 10^{24}$ atoms. Mass $= (1.8 	imes 10^{24} / 6 	imes 10^{23}) 	imes 235 = 3 	imes 235 = 705 	ext{ g} = 0.705 	ext{ kg}$.

Similar Questions

$A$ nuclear fission process is given by $A^{240} \rightarrow B^{100} + C^{140} + Q$ (energy). If the binding energy per nucleon for nuclei $A, B,$ and $C$ are $7.6 \, MeV, 8.1 \, MeV,$ and $8.1 \, MeV$ respectively,then the energy $Q$ released is approximately $...... \, MeV$.

$_1H^1 + _1H^1 + _1H^2 \to X + _1e^0 + \text{Energy}$. The emitted particle $X$ is ........?

Fusion reaction is initiated with the help of

When $_{92}U^{235}$ undergoes fission,$0.1\%$ of its original mass is converted into energy. How much energy is released if $1\,kg$ of $_{92}U^{235}$ undergoes fission?

Nuclear fission was discovered by

Vedclass Test Series

Exam Paper Generator

Online Exam Module