Write the initial and final nuclei

Initial nucleus: ${}_{92}^{238}U$

Final nucleus: ${}_{92}^{234}U$

Effect of one $\alpha$-particle emission

An $\alpha$-particle is: ${}_2^4\alpha$

So after one $\alpha$ emission:

• Mass number decreases by $4$
• Atomic number decreases by $2$

Therefore, ${}_{92}^{238}U \xrightarrow{\alpha} {}_{90}^{234}Th$

Compare with required final nucleus

After $\alpha$ emission we have: ${}_{90}^{234}Th$

But required final nucleus is: ${}_{92}^{234}U$

So now:

• Mass number is already correct: $234$
• Atomic number must increase from $90$ to $92$

Which emission increases atomic number by 1 without changing mass number?

In $\beta^-$ emission: $n \to p + e^- + \bar{\nu}$

Hence:

• Atomic number increases by $1$
• Mass number remains unchanged

Therefore, to increase atomic number from $90$ to $92$, we need two $\beta^-$ emissions.

Check the full sequence

${}_{92}^{238}U \xrightarrow{\alpha} {}_{90}^{234}Th$ ${}_{90}^{234}Th \xrightarrow{\beta^-} {}_{91}^{234}Pa$ ${}_{91}^{234}Pa \xrightarrow{\beta^-} {}_{92}^{234}U$

Net result: ${}_{92}^{238}U \to {}_{92}^{234}U + \alpha + 2\beta^-$

Evaluate options

• A: Two $\beta^-$ — Correct
• B: Two $\beta^-$ and one $\beta^+$ — $\beta^+$ lowers atomic number by $1$, so incorrect
• C: One $\beta^-$ and one $\gamma$ — $\gamma$ does not change atomic number; net increase only $1$, so incorrect
• D: One $\beta^+$ and one $\beta^-$ — net change in atomic number is $0$, so incorrect

Final answer

The other emissions should be: $\boxed{\text{Two } \beta^-}$