Thorium disintegration series is known as:

(4n + 2) series

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(4n + 3) series

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(4n) series

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(4n + 1) series

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Solution

The correct option is C (4n) series
$\begin{matrix} S e r i e s & 4 n & 4 n + 1 & 4 n + 2 & 4 n + 3 N a m e & T h o r i u m & N e p t u n i u m & U r a n i u m & A c t i n i u m P a r e n t e l e m e n t & _{90} T h^{232} & _{94} P u^{241} & _{92} U^{238} & _{92} U^{235} P r o m i n e n t e l e m e n t & _{90} T h^{232} & _{93} N p^{237} & _{92} U^{238} & _{89} A c^{222} E n d p r o d u c t & _{82} P b^{208} & _{83} B i^{209} & _{82} P b^{206} & _{82} P b^{207} N u m b e r o f & α = 6 & α = 8 & α = 8 & α = 7 p a r t i c l e s l o s t & β = 4 & β = 5 & β = 6 & β = 4 \end{matrix}$

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_{90} T h^{232}

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In the disintegration series In the disintegration series $_{92}^{238} U α \to X β^{-} - \to_{Z}^{A} Y$ The values of Z and A respectively will be

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