How many α and β-particles are emitted when ₉₀Th²³² changes to ₈₂Pb²⁰⁸?
Answers
Answered by
0
Answer:
Explanation:
no. of alpha particles=232-208/4
=24/4
=6
no. of beta particles = 2*alpha+(z-z1)
=12-8
=4
Answered by
1
First 90Th232 changes into 88Ra228 with the emission of one alpha particle.
The reaction goes on and finally 82Pb208 forms.
Total number of alpha particles emitted is 7 and beta 4.
Attachments:
Similar questions