Question

naturally occurring uranium-238 is used to produce fuels for nuclear reactors. as uranium-238 decays, it only emits radiation that cannot pass through a sheet of paper. select the nuclear reaction that describes this process. \\(\ce{_{92}^{238}u \
ightarrow _{-1}^0e + _{93}^{238}np}\\) \\(\ce{_{92}^{238}u \
ightarrow _2^4\alpha + _{90}^{234}th}\\) \\(\ce{_{92}^{238}u \
ightarrow _{91}^{234}pa + _2^4he + _{-1}^0\beta}\\) \\(\ce{_{92}^{238}u \
ightarrow _{90}^{234}th + _0^0\gamma + _2^4\alpha}\\)

Explanation:

Step1: Recall decay types

Alpha decay emits $\alpha$ (${}_2^4\text{He}$), beta emits ${}_{-1}^0\text{e}$, gamma is $\gamma$. Alpha particles can't pass through paper.

Step2: Analyze each option

• Option 1: Emits ${}_{-1}^0\text{e}$ (beta), beta can pass through paper.
• Option 2: Emits ${}_2^4\text{He}$ (alpha), alpha can't pass through paper. Check mass/charge: $^{238}_{92}\text{U} \to {}_2^4\text{He} + {}^{234}_{90}\text{Th}$. Mass: 238 = 4 + 234; Charge: 92 = 2 + 90. Balanced.
• Option 3: Emits beta and alpha, but we need the one with alpha (can't pass paper).
• Option 4: Emits gamma and alpha, but the key is alpha emission (as alpha can't pass paper), and option 2 is pure alpha decay.

Answer:

The second option (${}^{238}_{92}\text{U} \to {}_2^4\alpha + {}^{234}_{90}\text{Th}$)