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.92U^238 decays by emitting successively...

`._92U^238` decays by emitting successively 8 `alpha`-particles and 8 `beta`-particles. Determine the mass number and atomic number of the new element and express it in symbol.

Text Solution

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Using the Soddy-Fajan's displacement rule,
Loss in mass number due to `alpha`-emission = 8 x 4 =32 and there is no change in mass number due to `beta`-emission
Hence, the mass number of the element formed
=-238-32 =206
Due to emission of 8 `alpha`-particles the decrease in atomic number
= 8 x 2 =16
Now , due to emission of 6 `beta`-particles the increase in the atomic number
=6 x 1 =6
`therefore` The atomic number of new element
= 92 -(16-6) = 92-10 =82
Atomic number being 82, the element formed is lead (Pb) and the symbolic representation is `._92Pb^206` .
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