Question

When $T_{90}^{228}$ transforms to $Bi_{83}^{212}$, then the number of the emitted $\alpha$ and $\beta$ particles is respectively:
A. $8\alpha ,7\beta$
B. $4\alpha ,7\beta$
C.$4\alpha ,4\beta$
D.$4\alpha ,1\beta$

Answer 1

Here we will be using the law of conservation of energy and law of conservation of Atomic number (charge) to solve these types of questions.

Complete step by step answer:
We can represent the nuclear reaction where $T_{90}^{228}$ changes to $Bi_{83}^{212}$ as follows:
$T_{90}^{228} \to Bi_{83}^{212} + xHe_2^4 + y\beta _{ - 1}^0$
Let us consider x and y as the number of the $\alpha {\text{ and }}\beta$particles respectively emitted when $T_{90}^{228}$ transforms into $Bi_{83}^{212}$. Hence, we know that this reaction is a combination of alpha and beta decay.
There are three types of decay processes, namely, alpha decay, beta decay, and gamma decay.
$\alpha$- decay: When a radioactive nucleus emits an$\alpha$- particle(${\text{He}}_2^4$) we call the process as $\alpha$- decay. When a nucleus emits an alpha particle (${\text{He}}_2^4$), it loses two protons and two neutrons. Therefore, the atomic number(Z) decreases by 2 and the atomic mass(A) decreases by 4. Hence, we can represent$\alpha$- decay as:
$X_Z^A \to Y_{Z - 2}^{A - 4} + He_2^4 + energy$
Where x is called the parent nucleus and Y is called the daughter nucleus.
$\beta$-decay: The process of emission of an electron from a nucleus is called $\beta$-decay.
When a nucleus emits a $\beta$- particle (electron or positron), the mass number A of the nucleus does not change but the atomic number Z increases by 1. We represent $\beta$- decay as:
$X_Z^A \to Y_{Z + 1}^A + e_{ - 1}^0 + energy$
Let us consider the reaction given in the question: $T_{90}^{228} \to Bi_{83}^{212}$
Now, we can represent the complete decay process as following:
$T_{90}^{228} \to Bi_{83}^{212} + xHe_2^4 + y\beta _{ - 1}^0$
Let us find the values of X and Y.
The atomic number of thorium isotope in this reaction is 90. Similarly, we can see that the atomic number of bismuth is 83.
We will apply the law of conservation of atomic number in the reaction equation. We will get $90 = 83 + 2x - y$
Now, we will bring the numerical values together. We get, $90 - 83 = 2x - y$
When we subtract the values, the equation becomes, $2x - y = 7$
Applying law of conservation of mass number, we get,
$228 = 212 + 4x$
$16 = 4x$
$x = 4$
Substituting the value for X in equation 1 we get,
$2x - y = 7$
$2\left( 4 \right) - y = 7$
$y = 1$

$\therefore$ There are 4 $\alpha$-particles and $1{\text{ }}\beta$ -particles. Hence the correct option is (D).

Note:
For this type of question, we have to learn how to write the nuclear reaction equation for the given nuclear reaction. We must also study the atomic number and the atomic mass number of the most common radioactive isotopes.
We can solve this question with another alternate method. Let us represent the equation as follows:
$T_{Z = 90}^{A = 228} \to Bi_{{Z^1} = 83}^{{A^1} = 212}$
In this method, we will use the formula, ${n_\alpha } = \dfrac{{A - {A^1}}}{4}$
Number of alpha particles emitted,
${n_\alpha } = \dfrac{{A - {A^1}}}{4}$
$= \dfrac{{228 - 212}}{4}$
$= 4$
Number of beta particles emitted,
${n_\beta } = 2{n_\alpha } - Z + {Z^{^1}}$
${n_\beta } = 2 \times 4 - 90 + 83$
${n_\beta } = 1$