Question

A radioactive nucleus (initial mass number A and atomic number Z) emits 3$\alpha $- particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be
A.$\dfrac{{{\rm{A - Z - 8}}}}{{{\rm{Z - 4}}}}$
B. $\dfrac{{{\rm{A - Z - 4}}}}{{{\rm{Z - 8}}}}$
C. $\dfrac{{{\rm{A - Z - 12}}}}{{{\rm{Z - 4}}}}$
D. $\dfrac{{{\rm{A - Z - 4}}}}{{{\rm{Z - 2}}}}$

Answer 1

Hint:An alpha particle is a helium nucleus. When an alpha particle is emitted, the parent element's atomic number is reduced by 2, and the mass number is reduced by 4. When a beta particle is emitted, the atomic number increases by one, and the mass number remains unchanged.

The mass number A is the sum of the number of neutrons (n) and the number of protons (p).
${\rm{A}}\; = n + p$
The atomic number Z is equal to the number of electrons or number of protons
Therefore, ${\rm{Z}}\;{\rm{ = }}\;{\rm{p}}$

Complete step by step answer:
Radioactive decay is a process in which an unstable nucleus loses energy by radiation. It can be of three types:
- $\alpha $- decay
- $\beta $-decay
- $\gamma $- decay
An alpha particle is a helium nucleus. When an alpha particle is emitted, the parent element's atomic number is reduced by 2, and the mass number is reduced by 4. When a beta particle(positron) is emitted, the atomic number increases by one, and the mass number remains unchanged.

Suppose the parent nuclei are represented as ${\rm{X}}_{\rm{Z}}^{\rm{A}}$, where A is the mass number, and Z is the atomic number.
We know, (A= n+p)
The mass number A is the sum of the number of neutrons (n) and the number of protons (p).
The atomic number Z is equal to the number of electrons or number of protons
Therefore, ${\rm{Z}}\;{\rm{ = }}\;{\rm{p}}$ .
Now let us suppose an alpha particle is emitted from this nucleus. Then the daughter nuclei will be given by

Similarly, in beta emission, the atomic number is increased by one. Thus

When $3\alpha $-particles are emitted the atomic number of the parent element is reduced by$3 \times 2$ and the
The mass number is reduced by $3 \times 4$
That is, we get a new mass number = A-12…. (1)
 New atomic number = Z-6
Now, if $2\beta $-particles (positrons) are emitted,
The mass number remains the same, but the new atomic number will be $ = {\rm{Z - [6 + (2}} \times {\rm{1)]}}\;{\rm{ = }}\;Z - \;8$….. (2)
The formula can found out the number of neutrons (n),
 $ \Rightarrow \;n$= mass number – atomic number
Substitutes the value from 1and 2 equation we get,
$n = \,{\rm{A - 12 - ( Z - 8)}}$
$ \Rightarrow \;n\; = \;{\rm{A - Z - 4}}$
No of protons in the nucleus is $p\; = {\rm{Z - }}\;8$
Thus,
$\dfrac{n}{p} = \;\dfrac{{{\rm{A - Z - 4}}}}{{{\rm{Z - 8}}}}$
Therefore, the ratio of the number of neutrons and protons in the product nucleus is$\dfrac{{{\rm{A - Z - 4}}}}{{{\rm{Z - 8}}}}$.

Hence the correct option is (A).

Note:There are two types of beta decay. ${\beta ^ - }$ - decay and ${\beta ^ + }$-decay. In the parent nucleus, a ${\beta ^ - }$-decay neutron is converted into a proton, electron, and an electron antineutrino. Here electrons are emitted, and the atomic number of the parent nucleus is increased by 1. Whereas in ${\beta ^ + }$ decay, a proton in the parent nuclei is converted into a positron, neutron, and electron neutrino. Here positron is emitted, and the atomic number of the parent nuclei is decreased by 1.