In the given nuclear reaction $A, B, C, D, E$ represents

$_{92}{U^{238}}{\xrightarrow{\alpha }_B}T{h^A}{\xrightarrow{\beta }_D}P{a^C}{\xrightarrow{E}_{92}}{U^{234}}$

  • A

    $A = 234, B = 90, C = 234, D = 91, E =  \beta $

  • B

    $A = 234, B = 90, C = 238, D = 94, E = \alpha $

  • C

    $A = 238, B = 93, C = 234, D = 91, E = \beta $

  • D

    $A = 234, B = 90, C = 234, D = 93, E = \alpha $

Similar Questions

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