Get Answers to all your Questions

header-bg qa

If the energy released in the fission of one nucleus is 200 MeV. Then the number of nuclei required per second in a power plant of 16 kW will be

Option: 1

0.5\times 10^{14}


Option: 2

0.5 \times 10^{12}


Option: 3

5 \times 10^{12}


Option: 4

5 \times 10^{14}


Answers (1)

best_answer

Energy released in the fission of one nucleus =200 \, \mathrm{MeV}

=200 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}=3.2 \times 10^{-11} \mathrm{~J}

\mathrm{P=16 \mathrm{KW}=16 \times 10^3 \, \mathrm{Watt}}

Now, number of nuclei required per second

\mathrm{n=\frac{P}{E}=\frac{16 \times 10^3}{3.2 \times 10^{-11}}=5 \times 10^{14} }.
 

Posted by

Divya Prakash Singh

View full answer

JEE Main high-scoring chapters and topics

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