Question

A radioactive nucleus $X_Z^A$ emits $3\alpha $ particles and $5\beta $ particles. What will be the ratio of the number of neutrons and protons in the product nucleus?
A. $\dfrac{{A - Z - 12}}{{Z - 6}}$
B. $\dfrac{{A - Z}}{{Z - 1}}$
C. $\dfrac{{A - Z - 11}}{{Z - 1}}$
D. $\dfrac{{A - Z - 12}}{{Z - 1}}$

Answer 1

Hint: An alpha particle is a helium nucleus. When an alpha particle is emitted the atomic number of the parent element 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 $p$.
$A = n + p$
The atomic number $Z$ is equal to the number of electrons or the number of protons.
$Z = p$

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
1. $\alpha$ decay
2. $\beta$ decay
3. $\gamma$ decay
An alpha particle is a helium nucleus. When an alpha particle is emitted the atomic number of the parent element 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
Suppose the parent nuclei is represented as $X_Z^A$, where $A$ is the mass number and $Z$is the atomic number
The mass number $A$ is the sum of the number of neutrons $n$ and number of protons $p$
$A = n + p$
The atomic number $Z$ is equal to the number of electrons or number of protons
$Z = p$
Now let us suppose an alpha particle is emitted from this nucleus. Then the daughter nuclei will be given by
$X_Z^A\xrightarrow{\alpha }Y_{Z - 2}^{A - 4}$
Similarly, in beta emission, the atomic number is increased by one. Thus
$X_Z^A\xrightarrow{\beta }Y_{Z + 1}^A$
When $3\alpha $ particles are emitted the atomic number of the parent element is reduced by $3 \times 2$ and the mass number is reduced by $3 \times 4$
That is, we get new mass number \[ = A - 12\]............(1)
 new atomic number \[ = Z - 6\]
Now, if $5\beta $ particles are emitted,
The mass number remains the same but the new atomic number will be $Z - 6 + (5 \times 1)$
That is, Atomic number = $Z - 1$.......................(2)
The number of neutrons($n$) can be found out by the formula,
$\text{n = Mass number - Atomic number}$
Substitute the values from equation one and two in the above formula, Then we get
$ n = A - 12 - (Z - 1) $
$n = A - Z - 11 $
Number of protons in the nucleus is \[p = Z - 1\]
Thus,
$\dfrac{n}{p} = \dfrac{{A - Z - 11}}{{Z - 1}}$

Therefore, the ratio of the number of neutrons and protons in the product nucleus is $\dfrac{{A - Z - 11}}{{Z - 1}}$. Hence, the correct answer is option C.

Note:
There are two types of beta decay. ${\beta ^{^ - }}$decay and ${\beta ^{^ + }}$ decay. In ${\beta ^{^ - }}$ decay neutron in the parent nucleus 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 $1$. In this question, we considered the ${\beta ^{^ - }}$ decay where the atomic number of the parent nucleus increases by $1$.