Question

In the uranium radioactive series, the initial nucleus is $U_{92}^{238}$ and that the final nucleus is $Pb_{82}^{206}$. When uranium nucleus decays to lead the number of $\alpha$- particles and $\beta$- particles emitted are:
A) $8\alpha ,6\beta$
B) $6\alpha ,7\beta$
C) $6\alpha ,8\beta$
D) $4\alpha ,3\beta$

Answer 1

The radioactive decay occurs in series where daughter product gives rise to granddaughter and so on. Some of them are alpha emitters while others are beta emitters. We can find one of the particles by subtracting the atomic number from the parent nucleus and with the help of that particle, we can find the other.

Complete step by step solution:
Chemical equation:
$U_{92}^{238} \to Pb_{82}^{206} + {\text{xHe}}_2^4 + {\text{y}}\beta _{ - 1}^0$
Let x and y be the number of $\alpha {\text{ and }}\beta$particles respectively emitted when $U_{92}^{238}$ transforms into $Pb_{82}^{206}$. This reaction is combination of alpha and beta decay.
There are three types of Decay process:
(i) Alpha decay
(ii) Beta decay
(iii) Gamma decay
(i) The process of emission of $\alpha$- particle (${\text{He}}_2^4$) from a radioactive nucleus is called $\alpha$- decay. When a nucleus emits an alpha particle (${\text{He}}_2^4$), it loses two protons and two neutrons. Therefore atomic number(Z) decreases by 2 and atomic mass(A) decreases by 4. Hence $\alpha$- decay can be represented as:
${\text{X}}_Z^A \to {\text{Y}}_{Z - 2}^{A - 4} + {\text{He}}_2^4 + {\text{energy}}$
Where $x$ is called the parent nucleus and Y is called daughter nucleus.
(ii) 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 its atomic number Z increases by 1 .Therefore, $\beta$- decay can be represented as:
${\text{X}}_Z^A \to {\text{Y}}_{Z + 1}^A + e_{ - 1}^0 + {\text{energy}}$
Consider a given reaction, ${\text{T}}_{90}^{228} \to {\text{Bi}}_{83}^{212}$
Now, we can write the complete decay process in terms of chemical equation.
$U_{92}^{238} \to Pb_{82}^{206} + {\text{xHe}}_2^4 + {\text{y}}\beta _{ - 1}^0$
Let’s find the values of X and Y.
Applying law of conservation of atomic number, we have,
$\Rightarrow 92 = 82 + 2{\text{x}} - {\text{y}}$
$\Rightarrow 92 - 82 = 2{\text{x - y}}$
$\Rightarrow 2{\text{x}} - {\text{y}} = 10.......{\text{(1)}}$
Applying law of conservation of mass number, we get,
$\Rightarrow 238 = 206 + 4{\text{x}}$
$\Rightarrow 32 = 4{\text{x}}$
$\therefore {\text{x}} = 8$
Substituting the value of X in equation (1),
$\Rightarrow 2{\text{x}} - {\text{y}} = 10$
$\Rightarrow 2\left( 8 \right) - {\text{y}} = 10$
$\therefore {\text{y}} = 6$

Therefore, there are $8- \alpha$-particles and ${\text{6 }}\beta$ particles. So, the correct option is (A).

Note:
The process of spontaneous disintegration of the nucleus of a heavy element with the emission of some particle or electromagnetic radiation.
There is one more alternative method.
$U_{Z = 92}^{A = 238} \to Pb_{{Z^1} = 82}^{{A^1} = 206}$
Number of alpha particles emitted,
${n_\alpha } = \dfrac{{A - {A^1}}}{4}$
$\Rightarrow \dfrac{{238 - 206}}{4}$
$\therefore{n_\alpha } = 8$
Number of beta particles emitted,
${n_\beta } = 2{n_\alpha } - Z + {Z^{^1}}$
$\Rightarrow {n_\beta } = 2 \times 8 - 92 + 82$
$\therefore {n_\beta } = 6$