CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Assuming that about 200 MeV of energy is released per fission of 92U235 nuclei, then the mass of 92U235 consumed per day in fission reactor of power 1 MW will be approximately

A
102 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1 g
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
100 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
10 g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 1 g
Given,
Power of fission reactor is, P=106 W=106 Js1

Time required =t=1 day=24×36×102=8.64×104 s

Energy produced, U=Pt=106×8.64×104=8.64×1010 J

Energy released per fission of 92U235 is,

200 MeV=200×106×1.6×1019=32×1012 J

The number of 92U235 atoms used is,

=8.64×101032×1012=27×1020

The mass of 92U235 required is,

m=(2356×1023)(27×1020)=1.058 g1 g

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (B) is the correct answer.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Nuclear Fission
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon