Question

A radioactive nucleus ( initial mass number a and atomic number z) emits 3 alpha particles and 2 positrons the ratio of number of neutrons to that of protons in the final nucleus will be

Answer 1

Hello dear,

● Answer-
neutrons/protons = (A-Z-4)/(Z-8)

● Explaination-
When a radioactive nucleus emits 1 α-particle mass number decreases by 4 and atomic number decreases by 2.
A' = A-4×3 = A-12
Z' = Z-2×3 = Z-6

When this nucleus emits 1 β-particle, atomic mass remians unchanged but atomic number decreses by 1.
A" = A' = A-12
Z" = Z'-2×1 = Z-8

Now,
Protons = Z" = Z-8
Neutrons = A"-Z" = A-Z-4

Ratio of neutrons to protons-
n/p = (A-Z-4)/(Z-8)

Hope this is helpful...

Answer 2

X emits 3 alpha particles. One alpha particle emission results in reduction of mass number by 4 and atomic number by 2. So emission of 3 alpha particles results in reduction of mass number by 12 and atomic number by 6. It also emits 2 positrons which results in further reduction of the atomic number by 2.

So, in the final nucleus,

Number of protons = Z - 6 - 2 = Z - 8

Number of neutrons = mass number - atomic number = A - 12 - (Z - 8) = A - Z - 4

Therefore,

Number of neutrons / number of protons = A - Z - 4 / Z - 8

Thank u.

Hope this helps you.

Please mark as brainliest if you think it is so.

@spyder