Answer: The given decay sequence is [tex]^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta[/tex].
Explanation:
An alpha-particle is a helium atom. Hence, when an alpha decay occurs in [tex]^{238}_{92}U[/tex] then the reaction equation is as follows.
[tex]^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He[/tex]
Now, in sequence the equation for beta decay is as follows.
[tex]^{234}_{90}Th \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta[/tex]
Hence, the sequence will be as follows.
[tex]^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta[/tex]
Thus, we can conclude that the given decay sequence is [tex]^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta[/tex].
