Question

Which nuclide will decay by positron and what isotope will be the left over? Both answers must be correct. (Numbers represent the mass number): 32S, 32P; 26P, 26S; 14O, 14N; 14N, 14O

Answer 1

32P
26P
14O

This all things will decay due to positron emission

Explanation:

Laws of radioactivity

If `N/Z` is too high beta decay takes place
If `N/Z` is too low positron emission or alpha decay takes place

For atomic no. = 1 `rarr` 20
stable if `N/Z` = 1

For atomic no.= 20 `rarr` 40

stable if `N/Z` ratio 1.25 `rarr`1.5

For atomic n= 40 `rarr` 80
stable if #N/Z ratio 1.5

For atomic no. > 83
no stable nuclei

32S has 16 neutrons , 16 protons

atomic no. = 16
`therefore N/Z = 16/16 = 1`
so it will not decay

32P has 17neutrons and 15protons

atomic weight = 15
`therefore N/Z = 17/15 = 1.133("not equal to 1.5 ")`
From this we know that 32P is not stable but type of radioactive decay is positron emission as 1.133< 1.5
`""_32^15P→` `""_32^16Si^+` + `e^−` + νe

26P
protons = 15
Z(protons) = 15
N(neutrons) = 11
`N/Z = 11/15 = 0.7333("not 1")`
26P is not stable but as 0.733 is too low than 1 it is a positron emission
`""_15^26P→` `""_14^26Si^+` + `e^−` + νe

26S No. of protons = 16
No. of neutrons = 10
`N/Z = 10/16 = 0.625`
Its not stable but still its a proton emission not positron emission.
But as 0.625<1 this radioactive decay follows a positron emission

14N
No. of protons = 7
No. of neutrons = 7
`N/Z = 7/7 = 1`
It is stable

14O
No. of protons = 8
No. of neutrons = 6
`N/Z = 6/8 = 0.75`
Too low than 1 so positron emission

`""_8^14P→` `""_7^14N` +`e^−`