A) 4a and 2b
B) 6a and 4b
C) 8a and 24b
D) 4a and 16b
Correct Answer: B
Solution :
\[{{n}_{\alpha }}=\frac{A-A'}{4}=\frac{232-208}{4}=6\] \[{{n}_{\beta }}=2{{n}_{\alpha }}-Z+Z'=2\times 6-90+82=4\]You need to login to perform this action.
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