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

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1 Answer

Answer:

32P
26P
14O

This all things will decay due to positron emission

Explanation:

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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

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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^− #