Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The energy liberated on complete fission of 1 \mathrm{~kg} of \mathrm{{ }_{92} U^{235}} is (Assume \mathrm{200 \, \mathrm{MeV}} energy is liberated on fission of 1 nucleus)

Option: 1

8.2 \times 10^{10} \mathrm{~J}


Option: 2

8.2 \times 10^9 \mathrm{~J}


Option: 3

8.2 \times 10^{13} \mathrm{~J}


Option: 4

8.2 \times 10^{16} \mathrm{~J}


Answers (1)

best_answer

Mass of a uranium nucleus
\mathrm{=92 \times 1.6725 \times 10^{-27}+143 \times 1.6747 \times 10^{-27} }
\mathrm{=393.35 \times 10^{-27} \mathrm{~kg}}

Number of nuclei in the given mass
\mathrm{=\frac{1}{393.35 \times 10^{-27}}=2.542 \times 10^{24} }

\mathrm{\text { Energy released }=200 \times 2.542 \times 10^{24} \mathrm{MeV} }

\mathrm{=5.08 \times 10^{26} \mathrm{MeV}=8.135 \times 10^{13} \mathrm{~J}=8.2 \times 10^{13} \mathrm{~J} }
 

Posted by

Rishi

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024: NTA debars 39 candidates for 3 years on account of using unfair means
anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

Latest Question