You are given two nuclei ${}_{3}^{7}X and {}_{3}^{4}Y .$ Explain giving reasons, as to which one of the two nuclei is likely to be more stable?

Solution

In case of ${}_{3}^{7}X$ ,

Neutron to proton ratio is,

$\frac{neutron number}{proton number} = \frac{7 - 3}{3} = \frac{4}{3} = 1.33$

In case of ${}_{3}^{4}Y$ , neutron to proton ratio is given by,
$\frac{neutron number}{proton number} = \frac{4 - 3}{3} = \frac{1}{3} = 0.33$

For stability, this ratio has to be close to one.

Therefore, nucleus ${}_{3}^{7}X$ is more stable than the nucleus ${}_{3}^{4}Y .$

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You are given two nuclei ${}_{3}^{7}X and {}_{3}^{4}Y .$ Explain giving reasons, as to which one of the two nuclei is likely to be more stable?