Question

The element ${}_{90}^{232}Th$ belongs to 4n series which will be the end product of this series:

• A. ${}_{83}^{209}Bi$
• B. ${}_{82}^{208}Bi$
• C. ${}_{82}^{209}Bi$
• D. ${}_{82}^{207}Bi$

Answer 1

The correct option is B ${}_{82}^{208}Bi$

$\begin{array}{ccccc}Series& 4n& 4n+1& 4n+2& 4n+3\\ Name& Thorium& Neptunium& Uranium& Actinium\\ Parentelement& {}_{90}T{h}^{232}& {}_{94}P{u}^{241}& {}_{92}{U}^{238}& {}_{92}{U}^{235}\\ Prominentelement& {}_{90}T{h}^{232}& {}_{93}N{p}^{237}& {}_{92}{U}^{238}& {}_{89}A{c}^{222}\\ Endproduct& {}_{82}P{b}^{208}& {}_{83}B{i}^{209}& {}_{82}P{b}^{206}& {}_{82}P{b}^{207}\\ Numberof& \alpha =6& \alpha =8& \alpha =8& \alpha =7\\ particleslost& \beta =4& \beta =5& \beta =6& \beta =4\end{array}$