# 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

Feb 19, 2017

32P
26P
14O

This all things will decay due to positron emission

#### Explanation:

If $\frac{N}{Z}$ is too high beta decay takes place
If $\frac{N}{Z}$ is too low positron emission or alpha decay takes place

For atomic no. = 1 $\rightarrow$ 20
stable if $\frac{N}{Z}$ = 1

For atomic no.= 20 $\rightarrow$ 40

stable if $\frac{N}{Z}$ ratio 1.25 $\rightarrow$1.5

For atomic n= 40 $\rightarrow$ 80
stable if N/Z ratio 1.5

For atomic no. > 83
no stable nuclei

32S has 16 neutrons , 16 protons

atomic no. = 16
$\therefore \frac{N}{Z} = \frac{16}{16} = 1$
so it will not decay

32P has 17neutrons and 15protons

atomic weight = 15
$\therefore \frac{N}{Z} = \frac{17}{15} = 1.133 \left(\text{not equal to 1.5 }\right)$
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
$\frac{N}{Z} = \frac{11}{15} = 0.7333 \left(\text{not 1}\right)$
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
$\frac{N}{Z} = \frac{10}{16} = 0.625$
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
$\frac{N}{Z} = \frac{7}{7} = 1$
It is stable

14O
No. of protons = 8
No. of neutrons = 6
$\frac{N}{Z} = \frac{6}{8} = 0.75$
Too low than 1 so positron emission

""_8^14P→ ""_7^14N +e^− #