Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The fission properties of _239 ^94 Pu are very similar to those of _235 ^92 U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure _239 ^94 Pu undergo fission?

Q. 13.17  The fission properties of _{94}^{239}\textrm{Pu} are very similar to those of _{92}^{235}\textrm{U}. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure _{94}^{239}\textrm{Pu}  undergo fission?

Answers (1)

best_answer

Number of atoms present in 1 kg(w) of  _{94}^{239}\textrm{Pu} =n

\\n=\frac{w\times N_{A}}{mass\ number\ of\ Pu}\\ n=\frac{1000\times 6.023\times 10^{23}}{239}\\n=2.52\times 10^{24}

Energy per fission (E)=180 MeV

Total Energy released if all the atoms in 1 kg  _{94}^{239}\textrm{Pu} undergo fission = E \times n

=180 \times 2.52\times1024

=4.536\times1026 MeV

 

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 Accountancy Exam Analysis 2025: Paper ‘moderate', includes application-based questions
anam-khan

CBSE mulls allowing non-programmable, basic calculators for class 12 Accountancy exam
anam-khan

CBSE Class 12 Biology Exam Analysis 2025: Paper 'moderate, well-structured'; passing marks, students' feedback

Latest Question