Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

If 200 \, \mathrm{MeV} energy is released in the fission of a single \mathrm{U^{235}} nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (Given \mathrm{1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}})

Option: 1

3.125 \times 10^{13}


Option: 2

3.125 \times 10^{14}


Option: 3

3.125 \times 10^{15}


Option: 4

3.125 \times 10^{16}


Answers (1)

best_answer

\mathrm{P =n\left(\frac{E}{t}\right) \Rightarrow 1000=\frac{n \times 200 \times 10^6 \times 1.6 \times 10^{-19}}{t}}

\mathrm{\Rightarrow \frac{n}{t}=3.125 \times 10^{13}}.

Posted by

Nehul

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