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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

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`._(90)Th^(232) rarr ._(82)Pb^(208) + x ._(2)He^(4) + y ._(-1)e^(0)`
Remembering that the total atomic number as well as mass number must be equal to the two sides of the equation, we have
`208 + 4x + 0y = 232`
i.e., `4x = 24`
and `82 + 2x - y = 90`
or `2x - y = 8`
Solving Eqs. (i) and (ii), we get `x = 6` and `y = 4`.
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