A) \[\alpha =3,\ \beta =3\]
B) \[\alpha =6,\ \beta =4\]
C) \[\alpha =6,\ \beta =0\]
D) \[\alpha =4,\ \beta =6\]
Correct Answer: B
Solution :
\[{{n}_{\alpha }}=\frac{A-A'}{4}=\frac{232-208}{4}=6\] and \[{{n}_{\beta }}=(2{{n}_{\alpha }}-Z+Z')=(2\times 6-90+82)=4\]
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