浏览科目Chemistry — Semester B
Chemistry · Semester B TEKS 14A-14C
HardCalc

Aluminum-27 is bombarded with an alpha particle, producing phosphorus-30 and one additional particle. Write the balanced nuclear equation for this artificial transmutation, identifying the missing particle by conserving both mass number and atomic number.

A²⁷₁₃Al + ⁴₂He → ³⁰₁₅P + ¹₀n — the missing particle (mass number 1, atomic number 0) is a neutron. This matches the correct reasoning chain worked through step by step from the given data.
B²⁷₁₃Al + ⁴₂He → ³⁰₁₅P + ¹₁H, incorrectly identifying the missing particle as a proton with atomic number 1 instead of 0
C²⁷₁₃Al + ⁴₂He → ³⁰₁₅P + ⁰₋₁e, identifying the missing particle as a beta particle, which balances neither mass nor atomic number here. This looks superficially similar to the correct approach but swaps a key detail that changes the outcome entirely.
D²⁷₁₃Al + ⁴₂He → ³⁰₁₅P + ⁴₂He, using another alpha particle as the product despite one already being consumed as a reactant

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