Isotopic abundance from weighted average atomic mass

Steps:

• Denote lower mass number as 121 (integer below average) and higher as 123.
• Let abundance of higher isotope be $p\%$, so lower is $(100 - p)\%$.
• Set up equation: $\frac{121(100 - p) + 123p}{100} = 121.76$.
• Simplify: $121 + 0.02p = 121.76$, so $p = 38$.

Why B is correct:

• 38% satisfies the weighted average formula, yielding exactly 121.76 when lower isotope (more abundant) dominates.

Why the others are wrong:

• A: 12% pulls average too low (≈121.24), ignoring the 0.76 deviation.
• C: 62% makes higher isotope more abundant, contradicting the problem statement.
• D: 68% pushes average too high (≈122.36), exceeding 121.76.

Final answer: B