Nuclear Chemistry - Result Question 9
9. A plot of the number of neutrons (n) against the number of protons $(p)$ of stable nuclei exhibits upward deviation from linearity for atomic number, $Z>20$. For an unstable nucleus having $n / p$ ratio less than 1 , the possible mode(s) of decay is (are)
(2016 Adv.)
(a) $\beta^{-}$- decay $(\beta$ - emission)
(b) orbital or $K$-electron capture
(c) neutron emission
(d) $\beta^{+}$-decay (positron emission)
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Answer:
Correct Answer: 9. (b,d)
Solution:
- For the elements with atomic number $(Z)$ larger than 20 ,
Neutrons $(n)>$ Protons $(p)$; Thus, $n / p>1$
Thus, there is upward deviation from linearity.
If $ n < p $, Thus $ n / p < 1 $, then
(a) By $\beta^{-}$- decay, ${ } _0^{1} n \longrightarrow{ } _1^{1} p+{ } _{-1}^{0} e$ neutron changes to proton. Thus, $(n / p)$ ratio further decreases below 1 . Thus, this decay is not allowed.
(b) By orbital or $K$ - electron capture, ${ } _1^{1} p+{ } _{-1}^{0} e \longrightarrow{ } _0^{1} n$ proton changes to neutron, hence, $(n / p)$ ratio increases. Thus stability increases. Thus correct.
(c) Neutron emission further decreases $n / p$ ratio.
(d) By $\beta^{+}$-emission, ${ } _1^{1} p \longrightarrow{ } _0^{1} n+{ } _{+1}^{0} e$ proton changes to neutron. Hence, $n / p$ ratio increases. Thus correct.
Plot of the number of neutrons against the number of protons in stable nuclei (shown by dots).
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