$^{234}_{90}Th → ^{234}_{91}Pa + X$ In the nuclear reaction above, X is

Analyze the question: Thorium-234 changes into protactinium-234. We need to find the particle X that is emitted, using conservation of mass number and atomic number.

Concept: In any nuclear reaction:

• Total mass number (top) on the left = total on the right
• Total atomic number (bottom) on the left = total on the right

Solution:

Mass number: 234 = 234 + A, so A = 0

Atomic number: 90 = 91 + Z, so Z = 90 − 91 = −1

So X has mass number 0 and charge −1, which is $^0_{-1}e$, a beta particle (β⁻).

This makes sense: in beta decay, a neutron changes into a proton plus an electron (n → p + e⁻). The electron is ejected, so the atomic number increases by 1 (90 → 91) while the mass number stays the same.

Checking the options:

• A. $^0_1e$ is a positron, which would give Z = 89, so it is wrong.
• B. $^0_{-1}e$ is a beta particle (electron). Correct.
• C. $^4_2He$ is an alpha particle, which would change the mass number, so it is wrong.
• D. $^1_0n$ is a neutron, which would change the mass number, so it is wrong.
• E. $^1_1p$ is a proton, which would change the mass number, so it is wrong.

Tip: If the mass number stays the same but the atomic number goes up by 1, it's beta (β⁻) decay. If the atomic number goes down by 1 with the same mass number, it's positron emission. If the mass number drops by 4 and the atomic number by 2, it's alpha decay.