90234Th disintegrates to give 82206Pb as the final product- Number of - and -particles are emitted out during this process is -

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Standard XII

Chemistry

Gamma Decay

90234Th disin...

Question

$_{90}^{234} Th$ disintegrates to give $_{82}^{206} Pb$ as the final product. Number of $α$ and $β$ -particles are emitted out during this process is :

[5 α, 7 β]

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[7 α, 6 β]

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[7 α, 5 β]

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none of the above

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Solution

The correct option is B $[7 α, 6 β]$
$α -$ particle - $_{2}^{4} H e$
$β -$ particle- $_{- 1}^{0} e$
$_{90}^{234} T h \to_{82}^{206} P b + a_{2}^{4} H e + b_{- 1}^{0} e$
$90 + 2 a - b = 82$
$234 - 4 a = 206$
$a = 7, b = 6$

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(IIT-JEE, 2004)

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23490Th disintegrates to give 20682Pb as the final product. Number of α and β-particles are emitted out during this process is :

The correct option is B [7α,6β]α− particle - 42Heβ− particle- 0−1e23490Th→20682Pb+a42He+b0−1e90+2a−b=82234−4a=206a=7,b=6

$_{90}^{234} Th$ disintegrates to give $_{82}^{206} Pb$ as the final product. Number of $α$ and $β$ -particles are emitted out during this process is :

The correct option is B $[7 α, 6 β]$
$α -$ particle - $_{2}^{4} H e$
$β -$ particle- $_{- 1}^{0} e$
$_{90}^{234} T h \to_{82}^{206} P b + a_{2}^{4} H e + b_{- 1}^{0} e$
$90 + 2 a - b = 82$
$234 - 4 a = 206$
$a = 7, b = 6$