Question

238 U 92 write the following decay sequence.​

Answer 1

Answer: The given decay sequence is $^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta$.

Explanation:

An alpha-particle is a helium atom. Hence, when an alpha decay occurs in $^{238}_{92}U$ then the reaction equation is as follows.

$^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He$

Now, in sequence the equation for beta decay is as follows.

$^{234}_{90}Th \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta$

Hence, the sequence will be as follows.

$^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta$

Thus, we can conclude that the given decay sequence is $^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He \rightarrow ^{234}_{91}Pa + ^{0}_{-1}\beta$.