The triad of nuclei that is isotonic is:

_{6}^{14} C,_{7}^{15} N,_{9}^{17} F

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_{6}^{12} C,_{7}^{14} N,_{9}^{19} F

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_{6}^{14} C,_{7}^{14} N,_{9}^{17} F

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_{6}^{14} C,_{7}^{14} N,_{9}^{19} F

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Solution

The correct option is A $_{6}^{14} C,_{7}^{15} N,_{9}^{17} F$
Isotonic : same no. of neutrons= mass number - atomic number

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Similar questions

_{6}^{14} C \to_{7}^{14} N + X

_{7}^{14} N +_{0}^{1} n ⟶_{6}^{14} C +_{1}^{1} H

Which of the following are isobars?

$_{18}^{40} A r$ $_{6}^{14} C$ $_{20}^{40} C a$ $_{7}^{14} N$

_{6}^{14} C \to_{7}^{14} N + X

β

_{6}^{14} C ⟶_{7}^{14} N +_{- 1}^{0} e

_{o}^{1} n ⟶_{1}^{1} p +_{- 1}^{o} e

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