Question

in alpha decay, alpha particles are ejected from the nucleus. which equation represents alpha decay?
\\( \ce{_{95}^{241}am -> _{93}^{237}np + _2^4he} \\)
\\( \ce{_{9}^{18}f -> _{8}^{18}o + _1^0e} \\)
\\( \ce{_{6}^{14}c -> _{7}^{14}n + _{-1}^0e} \\)
\\( \ce{_{66}^{152}dy -> _{66}^{152}dy + y} \\)

Explanation:

1. Recall the definition of alpha decay: In alpha decay, an atomic nucleus emits an alpha particle (which is a helium nucleus, $\ce{_{2}^{4}He}$), resulting in a new nucleus with a mass number reduced by 4 and an atomic number reduced by 2.
2. Analyze each option:

• Option A: The mass number of Am is 241, and after decay, the mass number of Np is 237 (241 - 4 = 237), and the atomic number of Am is 95, and after decay, the atomic number of Np is 93 (95 - 2 = 93), and it emits $\ce{_{2}^{4}He}$, which fits the alpha decay pattern.
• Option B: This emits a positron ($\ce{_{1}^{0}e}$), which is beta - plus decay, not alpha decay.
• Option C: This emits an electron ($\ce{_{- 1}^{0}e}$), which is beta - minus decay, not alpha decay.
• Option D: The nuclide remains the same and emits a gamma ray (represented by $y$), which is gamma decay, not alpha decay.

Answer:

A. $\ce{_{95}^{241}Am
ightarrow _{93}^{237}Np + _{2}^{4}He}$