Question

A radioisotope decays to give an alpha particle andPb-208.

Answer 1

Answer: The original element is $_{84}^{212}\textrm{Po}$

Explanation:

Alpha decay is defined as the process in which alpha particle is emitted. In this process, a heavier nuclei decays into a lighter nuclei. The alpha particle released carries a charge of +2 units.

The released alpha particle is also known as helium nucleus.

$_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha$

For the given alpha decay process of an isotope:

$^{Z}_{A}\textrm{X}\rightarrow ^{208}_{82}\textrm{Pb}+_2^4\alpha$

To calculate A:

Total mass on reactant side = total mass on product side

A = 208 + 4

A = 212

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

Z = 82 + 2

Z = 84

The isotopic symbol of unknown element is $_{84}^{212}\textrm{Po}$

Hence, the original element is $_{84}^{212}\textrm{Po}$