Use the standard changes in radioactive emissions

• For an $\alpha$-particle emission: $A \to A-4, \quad Z \to Z-2$
• For a $\beta^-$-particle emission: $A \to A, \quad Z \to Z+1$
• For a $\gamma$-ray emission: $A \to A, \quad Z \to Z$

Analyze the given decay sequence qualitatively

From the decay chain, the emitted species are labeled $x_1, x_2, x_3, x_4$.

From the nuclear changes shown in the sequence:

• The first step decreases mass number by $4$ and atomic number by $2$, so $x_1 = \alpha$
• The second step keeps mass number same and increases atomic number by $1$, so $x_2 = \beta^-$
• The third step does not correspond to a pure $\gamma$ transition, so $x_3 \neq \gamma$

Check each option

Option A: $Z$ is an isotope of uranium

Uranium has atomic number $92$. From the final nucleus shown in the sequence, $Z$ has atomic number $92$. Hence this is correct.

Option B: $x_2$ is $\beta^-$

As seen above, the second transition has: $\Delta A = 0, \quad \Delta Z = +1$ which is characteristic of $\beta^-$ emission. Hence correct.

Option C: $x_1$ will deflect towards negatively charged plate

Since $x_1 = \alpha$, it carries charge $+2$. A positively charged particle deflects toward the negatively charged plate. Hence correct.

Option D: $x_3$ is $\gamma$-ray

$x_3$ is not a $\gamma$ transition from the nuclear changes in the sequence. Hence incorrect.

Final answer

The correct options are: $\boxed{A,\ B,\ C}$

Comparison with stored correct answer

Stored correct answer: $A, B, C$

This matches the derived answer.