JEE Advanced 2016 · Paper 1 · Chemistry Q25

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ChemistryMultiple correct+4 / -2Question 25 of 54

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)

\beta^-

-decay (

\beta

emission)

orbital or

K

-electron capture

neutron emission

\beta^+

-decay (positron emission)

Answer and worked solution

Official answer: B, D

Proton-rich nucleus. With $N/P<1$ , the nucleus has too many protons. It moves towards stability by turning a proton into a neutron, either by:

$\beta^+$ emission: $p\to n+e^++\nu$ , or
orbital (K) electron capture: $p+e^-\to n+\nu$ .

Ruled out. β⁻ decay and neutron emission both lower $N/P$ further.

(B), (D)

Topics:Radioactive decay kinetics

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JEE Advanced 2016 · Paper 1, Chemistry question 25