Question

The atomic mass of thorium is 232.18 and its atomic number is 90. In terms of its radioactive disintegration, $6\alpha \text{ and 4}\beta $ particles are emitted. The mass number and atomic number of the product is:
[A] 208, 82
[B] 192, 78
[C] 212, 86
[D] 180, 72

Answer 1

Hint: The product we obtain is the one that we get on the disintegration of atoms of the 4n series. Emission of alpha particles decreases atomic number by 2 and mass number by 4. Emission of beta particles increases atomic number by 1.

Complete answer:
In radiochemistry, when a reaction takes place new elements are formed unlike chemical reaction where no new elements are formed.
We know that the charge of an alpha-particle is +2 electrons whereas the charge of a beta-particle is -1electron. We also know that the mass of an alpha particle is four times the mass of a hydrogen atom and the mass of a beta-particle is equal to the mass of an electron.
In the question which is given to us, 6 alpha and 4 beta-particles are emitted from thorium.
When an alpha-particle is emitted from an atom, the atomic number decreases by 2 and the mass number decreases by 4 and when a beta-particle is emitted, the atomic number increases by 1 and the mass number remains the same.
Here, 6 alpha particles are emitted which means the atomic number will decrease by 12 and the mass number will decrease by 24 also 4 beta particles are emitted which means atomic number will increase by 6 and mass number will remain the same.
Therefore, we can write that the net change in atomic number will be a decrease by 6 and the mass number decreases by 24.
In the question, the mass number of thorium is given which is 232.18 and the atomic number is 90. Therefore, the mass number of the atom after emission of 6 alpha and 4 beta particles will be (232-24) i.e. 208 and the atomic number will be (90-6) i.e. 82.
We can write the reaction as-
     \[{}_{90}^{232}Th-6\alpha -4\beta \to {}_{82}^{208}X\]
As we can see, the mass number and atomic number of the product is 208 and 82.

Therefore, the correct answer is option [A] 208, 82.

Note:
In the radioactive disintegration series, the first is 4n series which is the thorium series where 6-alpha and 4-beta particle releases from thorium and gives us an isotope of lead, which is \[{}_{82}^{208}Pb\]. In the above disintegration, the product we obtained is \[{}_{82}^{208}Pb\]. Similarly, there is Uranium series which is (4n+2) series where emission of 8-alpha and 6-beta particles from uranium-238 gives us \[{}_{82}^{206}Pb\] and we also have (4n+1) and (4n+3) series named Neptunium and Actinium series respectively. We get these series by dividing the mass number by 4. If it is completely divisible, the atom will undergo 4n series disintegration but if the remainder is 1, 2 or 3, it will be in (4n+1), (4n+2) or (4n+3) series respectively.