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In the given radioactive disintegration ...

In the given radioactive disintegration series,
`._90^(232)Th to _(2)^(208)Pb`
Calculate value of `(n+2)`.
Where value of n is number of isobars formed in this series, suppose there is successive emission of `beta-`particles.

Text Solution

Verified by Experts

`underset("Parent")(._(90)Th^(234) rarr ._(84)Po^(218)`
Decrease in mass `= (234 - 218) = 16`
Mass of 1 `alpha-"particle" `= 4 amu
Therefore, number of `alpha`-particles emitted
Number of `beta`-emitted - (Atomic number `= 2 xx` Number of `beta`-particles - Atomic number of end product)
`2 xx 4 - (90 - 84) (8 - 6) = 2`
Hence, number of `alpha`-particles = 4
and number of `beta`-particles = 2
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