Of the following nuclides, the one most likely, to be radioactive is:

_{15}^{31} P

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_{30}^{66} Z n

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_{7}^{14} N

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_{6}^{14} C

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Solution

The correct option is C $_{6}^{14} C$
$_{6}^{14} C$ is a radioactive element which is used in carbon-dating as its atomic number is below 20 and n/p ratio is greater than 1 as value is higher compared to P.

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Similar questions

(β^{+})

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

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

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

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