In the given radioactive disintegration series-23290Th - 20882 PbCalculate the value of n-2 where value of n is number of isobars formed in this series emission of - - particles-

^232_90Th → ^208_82Pb As mass is also changing along with atomic number it means α particles nbsp; are also emitted. No of nbsp;α particles emitted nbs ...

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Standard XII

Chemistry

Radioactive Series

In the given ...

Question

In the given radioactive disintegration series,
$_{90}^{232} T h \to_{82}^{208} P b$
Calculate the value of (n+2) where value of n is number of isobars formed in this series emission of $β$ − particles.

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Solution

$_{90}^{232} T h \to_{82}^{208} P b$

As mass is also changing along with atomic number it means $α$ particles are also emitted.

No of $α$ particles emitted
$1 α$ = $4 a m u$
$6 α$ = $24 a m u$
So atomic number must be decrease by 12 but it is decreasing only by 8
So $β$ particles emitted = 4
$I s o b a r s$ = nuclei having different atomic number but same mass number.
So, if mass number has to be same then it must be $β$ emission.
So, $5$ isobars are possible in total.

$A β ⟶ B β ⟶ C β ⟶ D β ⟶ E$

$n + 2$ = 7

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In the given radioactive disintegration series, 23290Th→20882Pb Calculate the value of (n+2) where value of n is number of isobars formed in this series emission of β − particles.

In the given radioactive disintegration series,
$_{90}^{232} T h \to_{82}^{208} P b$
Calculate the value of (n+2) where value of n is number of isobars formed in this series emission of $β$ − particles.