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lidiya [134]
3 years ago
6

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

Chemistry
1 answer:
oee [108]3 years ago
3 0

<u>Answer:</u> The original element is _{84}^{212}\textrm{Po}

<u>Explanation:</u>

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

<u>To calculate A:</u>

Total mass on reactant side = total mass on product side

A = 208 + 4

A = 212

<u>To calculate Z:</u>

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}

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