A) \[_{19}^{41}K\]
B) \[_{21}^{43}K\]
C) \[_{21}^{40}Sc\]
D) \[_{20}^{42}Ca\]
Correct Answer: C
Solution :
\[_{18}^{40}Ar\] having \[40 - 18 = 22\] neutrons While \[_{21}^{40}Sc\] having \[40 - 21 = 19\] neutrons.
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