When 3Li7 nuclei are bombarded by protons, and the resultant nuclei are 4Be8 , the emitted particles will be.
A
alpha particles
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
beta particles
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
gamma photons
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
neutrons
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C gamma photons 3Li7+11P⇒4Be8+00γ(energy) As, the emitted particles is only energy and gamma photons radiation is nothing but energy so, option C is correct.