QJEEAdv16Ch0981: JEE Advanced 2016 Chemistry - Class 12 JEEAdv16Ch0981 Solution: A plot of the number of neutrons (N) against the number of…

Plot of Z (number of protons) verses N (number of neutrons) is represented as :


(a) For a stable nucleus $\mathrm{N} / \mathrm{Z}\left(\frac{\text { no. of neutrons }}{\text { no. of protons }}\right)$ should be equal to 1.

This happens for elements till atomic number $(Z=20)$ . For such nucleus number of proton are equal to number of neutrons.

(b) For heavier nucleides with atomic number $(\mathrm{Z})$ greater than 20(Z>20) , the number of neutron are more than proton as seen by curved (green) appearance in the plot.

(c) For nucleids with \mathrm{N} / \mathrm{Z}<1 , the number of neutrons are less than the protons making the nucleus unstable. This is due to increase in coulombic repulsions between positively charged protons.

Such nuclei become stable by K-electron capture and positron emission.

K-electron capture :

$${}_1^{1}p+{}_{-1}^{0}e\to {}_0^{1}n$$

The K-electron capture, proton is converted to the neutron as increases the $\mathrm{N} / \mathrm{Z}$ ratio.

$\beta^{ \pm}$ decay (positron emission) :

$${}_1^{1}p\to \underset{neutron}{{}_0^{1}n}+\underset{positron}{e^{+}}+\underset{electronnutrino}{v_e}$$

Proton decays to a neutron, positron and electron nutrino. This decreases the number of protons and increases the number of neutrons, thereby increasing $\mathrm{N} / \mathrm{Z}$ ratio.

Options (B) and (D) are correct.