For a radioactive chain decay starting from 226-88 Ra and ending at 82206 PbA- Number of - decays is 4B- Number of - decays is 5C- Number of -decays is 4D- Number of -decays is 4

The correct options are B Number of α decays is 5 C Number of β^ decays is 4_88^226Ra→ _82^206Pb +n_1 _2^4He+n_2 β^ 88 = 82 + 2n_1 n_2 nbsp; nbsp; ...

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Byju's Answer

Standard XII

Chemistry

Half and Average Life

For a radioac...

Question

For a radioactive chain decay starting from $_{88}^{226} Ra$ and ending at $_{82}^{206} Pb$

Number of

α

decays is 4

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Number of

α

decays is 5

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Number of

β^{-}

decays is 4

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Number of

β^{+}

decays is 4

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Solution

The correct options are
B Number of $α$ decays is 5
C Number of $β^{-}$ decays is 4
$_{88}^{226} Ra \to_{82}^{206} Pb + n_{1}_{2}^{4} H e + n_{2} β^{-}$
$88 = 82 + 2 n_{1} - n_{2}$ ...(1)
$226 = 206 + 4 n_{1}$ ...(2)

From (2), $n_{1} = 5$
and From (1), $n_{2} = 4$

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For a radioactive chain decay starting from 22688Ra and ending at 20682Pb

The correct options are B Number of α decays is 5 C Number of β− decays is 422688Ra→ 20682Pb+n1 42He+n2 β− 88=82+2n1−n2 ...(1) 226=206+4n1 ...(2) From (2), n1=5 and From (1), n2=4

For a radioactive chain decay starting from $_{88}^{226} Ra$ and ending at $_{82}^{206} Pb$

The correct options are
B Number of $α$ decays is 5
C Number of $β^{-}$ decays is 4
$_{88}^{226} Ra \to_{82}^{206} Pb + n_{1}_{2}^{4} H e + n_{2} β^{-}$
$88 = 82 + 2 n_{1} - n_{2}$ ...(1)
$226 = 206 + 4 n_{1}$ ...(2)

From (2), $n_{1} = 5$
and From (1), $n_{2} = 4$