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 = 234, B = 90, C = 234, D = 91, E = \beta $
$A = 234, B = 90, C = 238, D = 94, E = \alpha $
$A = 238, B = 93, C = 234, D = 91, E = \beta $
$A = 234, B = 90, C = 234, D = 93, E = \alpha $
Consider the following radioactive decay process
${ }_{84}^{218} A \stackrel{\alpha}{\longrightarrow} A_1 \stackrel{\beta^{-}}{\longrightarrow} A_2 \stackrel{\gamma}{\longrightarrow} A_3 \stackrel{\alpha}{\longrightarrow} A_4 \stackrel{B^{+}}{\longrightarrow} A_5 \stackrel{\gamma}{\longrightarrow} A_6$
The mass number and the atomic number $A _6$ are given by
An atom of mass number $15$ and atomic number $7$ captures an $\alpha - $ particle and then emits a proton. The mass number and atomic number of the resulting product will respectively be
The radioactive nucleus $_7{N^{13}}$decays to $_6{C^{13}}$ through the emission of
When a radioactive substance emits an $\alpha$- particle, its position in the periodic table is lowered by
$\gamma$-decay occurs when