Z" = Z′ − 1 × 2 = Z − 6 − 2 = Z − 8
So, the ratio of neutrons to protons
n/p = (A − Z) − 12 / (Z − 8)

" /> Z" = Z′ − 1 × 2 = Z − 6 − 2 = Z − 8
So, the ratio of neutrons to protons
n/p = (A − Z) − 12 / (Z − 8)

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Question:

A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of the number of neutrons to that of protons in the final nucleus will be

A−Z−8Z−12

A−Z−12Z−8

A−Z−8Z−12

A−Z−102Z−8

Solution:

When a radioactive nucleus emits 1 α-particle mass number decreases by 4 and the atomic number decreases by 2.So after the emission of 3 α-particles
New atomic mass, A′ = A − 4 × 3 = A − 12
New atomic number Z′ = Z − 2 × 3 = Z − 6
When this nucleus emits 1 β-particle (positron), the atomic mass remains unchanged but the atomic number decreases by 1.So after emission of 2 positrons
A" = A′ = A − 12
Z" = Z′ − 1 × 2 = Z − 6 − 2 = Z − 8
So, the ratio of neutrons to protons
n/p = (A − Z) − 12 / (Z − 8)