Question

Atomic mass number of an element is 232 and its atomic number is 90. The end product of this radioactive element is an isotope of lead (atomic mass 208 and atomic mass 82). The number of $\alpha \text{ and }\beta \text{-}$ particles emitted is:
$\begin{align}
  & A.\text{ }\alpha \text{=3, }\beta \text{=3} \\
 & \text{B}\text{. }\alpha \text{=6, }\beta \text{=4} \\
 & \text{C}\text{. }\alpha =6,\text{ }\beta =0 \\
 & D.\text{ }\alpha \text{=4, }\beta \text{=6} \\
\end{align}$

Answer 1

Hint: We have been provided with atomic mass number and atomic number. To calculate $\alpha -\text{particle}$(number) write an equation such that it will give a relation between atomic mass number of radioactive element and atomic mass number of lead, according to mass balance similarly for $\beta -$ particle and hence calculate number of particles.

Complete answer:
In this question, we have been provided with radioactive elements whose atomic mass number is given 232 and whose atomic number is 90. We have another end product whose atomic mass is 208 and atomic number is 82. Now we need to find out the number of $\alpha \text{ and }\beta \text{-}$particles which are emitted from radioactive elements.
Let x be the number of particles i.e. $\alpha -\text{particle}$and y be the number of $\beta -$particles which are emitted. Now apply conservation of mass number. So initially if we conserve mass then it is given as 232 and finally is 208. We know that our $\alpha -\text{particle}$, if one $\alpha -\text{particle}$is emitted then atomic mass number is decreases by 4, so mathematically we can write it as,
$\begin{align}
  & 232-4x=208 \\
 & -4x=208-232 \\
 & x=\dfrac{232-208}{4}=\dfrac{24}{4} \\
 & x=6 \\
\end{align}$
This implies that, number of $\alpha -\text{particle}$emitted from radioactive elements is 6.
Consider similarly, for atomic numbers. We know that one $\alpha -\text{particle}$emits, atomic number decreases by two. So mathematically we can write it as,
$\begin{align}
  & 90-2x=82 \\
 & 90-2x=82......(1) \\
\end{align}$
But the same time, $\beta -$particles is also emitted by one, hence equation (1) can be written as,
$90-2x+y=82$
Put value x=6, we get
$\begin{align}
  & 90-2\times 6+y=82 \\
 & y=82-90+12 \\
 & y=4 \\
\end{align}$
This implies that, number of $\beta -$particles emitted from radioactive elements is 4. Hence the number of $\alpha \text{ and }\beta \text{-}$ particles emitted are 6 and 4.

Therefore the correct option is (b).

Note:
Students should know that alpha decay is a nuclear process in which an alpha particle is spontaneously ejected from the nucleus, whose mass number is greater than or equal to 140.
$\alpha -$ Decay process can be written as,
$\begin{align}
  & X_{Z}^{A}\xrightarrow{\alpha -decay}x_{Z-2}^{A-4}+{{H}_{2}}{{e}^{4}}+Q \\
 & \left( \text{parent nucleus} \right)\text{ }\left( \text{daughter nucleus} \right)\text{ }\left( \text{alpha-particle} \right) \\
\end{align}$
$\beta -$Decay process can be written as,
$\begin{align}
  & X{}_{Z}^{A}\xrightarrow{\beta -decay}Y_{Z+1}^{A}+\overline{\beta }+\overline{\gamma } \\
 & \left( \text{parent nucleus} \right)\left( \text{daughter nucleus} \right)\left( \text{beta particle} \right) \\
\end{align}$
Note that, Above question can also be solved as follows,
$_{90}T{{h}^{232}}{{\to }_{82}}p{{b}^{208}}+{{n}_{1}}\alpha +{{n}_{2}}{{\beta }^{-}}$
According to mass balance
$232=208+4\left( {{n}_{1}} \right)$
(Mass of alpha particle is 4)
${{n}_{1}}=6$
Now, v atomic number or proton balance according to
$\begin{align}
  & 90=82+2{{n}_{1}}-{{n}_{2}} \\
 & {{n}_{2}}=4 \\
\end{align}$