Question

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

What was the original element?
A.

B.
C.

D.

E.

Answer 1

Answer: The original element is
`_(84)^(212)\textrm{Po}`

Step-by-step 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}`