The element $_{90} T h^{232}$ belongs to thorium series. Which of the following will act as the end product of the series

_{82} P b^{208}

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_{82} B i^{209}

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_{82} P b^{206}

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_{82} P b^{207}

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Solution

The correct option is A $_{82} P b^{208}$
$\begin{matrix} S e r i e s & N a m e o f t h e s e r i e s & P a r e n t e l e m e n t & E n d s t a b l e e l e m e n t 4 n & T h o r i u m s e r i e s & T h - 232 & P b - 208 4 n + 1 & N e p t u n i u m s e r i e s & P u - 241 & B i - 209 4 n + 2 & U r a n i u m s e r i e s & U - 238 & P b - 206 4 n + 3 & A c t i n i u m s e r i e s & U - 235 & P b - 207 \end{matrix}$

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