A) 3a particles and 2b particle
B) 4a particles and 3b particle
C) 6a particles and 4b particle
D) 7a particles and 4b particle
Correct Answer: D
Solution :
\[{{n}_{\alpha }}=\frac{A-A'}{4}=\frac{235-207}{4}=7\] \[{{n}_{\beta }}=(2{{n}_{\alpha }}-Z+Z')=(2\times 7-92+82)=4\]
You need to login to perform this action.
You will be redirected in
3 sec
