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

A nucleus of mass number 220 decays by α decay. The energy released in the reaction is 5 MeV. The kinetic energy of an α-particle is:

A
154MeV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2711MeV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
5411MeV
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
5554MeV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 5411MeV
The decay proceeds as 220X216Y+4α
Since the momentum after the decay is conserved,
mαvα=mYvY
Thus vαvY=mYmα=2164=54
The ratio of their kinetic energies=KEYKEα=12mYv2Y12mαv2α=154
Q=KEY+KEα=5MeV
5554KEα=5MeV
KEα=5411MeV

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Rate of Change
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon