The number of alpha and beta ^- emitted durin | vedclass

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Question

Let no of $\alpha$ particle emitted $= x$ no of $\beta$ panticle emitted $=y$

Now, ${ }_{88}^{226} Ra \rightarrow{ }_{82}^{206} Pb + x _2^4 \alpha+

Vedclass · Accepted Answer

$5\alpha \ $ and $ \  4\beta ^-$

Vedclass · Answer

Let no of $\alpha$ particle emitted $= x$ no of $\beta$ panticle emitted $=y$

Now, ${ }_{88}^{226} Ra \rightarrow{ }_{82}^{206} Pb + x _2^4 \alpha+ y _{-1}^0 \beta$

balancing atomic mass

$226 =206+4 x+0$

$4 x =20$

$x =5$

balancing atomic no.

$88 =82+2 x+(-1) y$

$6 =10 \cdots y$

$y =4$

So, $5 \alpha$ and $4 \beta^{-}$

The number of $\alpha$ and $\beta ^-$ emitted during the radioactive decay chain starting from ${}_{88}^{226}Ra$ and ending at ${}_{82}^{206}Pb$ is

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