wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A nuclear fission is given below
A240B100+C140+Q(energy).
Let binding energy per nucleon of nuclei A,B and C be 7.6 MeV,8.1 MeV and 8.1 MeV respectively. Value of Q is : (Approximately)

A
20 MeV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
220 MeV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
120 MeV
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
240MeV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 120 MeV
Binding Energy of A=240×7.6=1824 MeV
Binding Energy of B=100×8.1=810 MeV
Binding Energy of C=140×8.1=1134 MeV
So Q=(810+1134)MeV1824 MeV=120 MeV

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