In alpha decay, alpha particles are ejected from the nucleus. Which equation represents alpha decay?

A. [tex]{}_{95}^{241} \text{Am} \rightarrow {}_{93}^{237} \text{Np} + {}_2^4 \text{He}[/tex]

B. [tex]{}_{9}^{18} \text{F} \rightarrow {}_{8}^{18} \text{O} + {}_{1}^{0} \text{e}[/tex]

C. [tex]{}_{6}^{14} \text{C} \rightarrow {}_{7}^{14} \text{N} + {}_{-1}^{0} \text{e}[/tex]

D. [tex]{}_{66}^{152} \text{Dy} \rightarrow {}_{66}^{152} \text{Dy} + \gamma[/tex]



Answer :

Alpha decay is a type of radioactive decay in which an unstable atomic nucleus ejects an alpha particle (which is essentially a helium-4 nucleus, [tex]\( {}_2^4 \text{He} \)[/tex]). In alpha decay, the parent nucleus loses 2 protons and 2 neutrons, resulting in a daughter nucleus with a mass number decreased by 4 and an atomic number decreased by 2.

Given this behavior of alpha decay, we need to identify which equation correctly represents this process:

1. [tex]\({ }_{95}^{241} \text{Am} \rightarrow { }_{93}^{237} \text{Np} + { }_{2}^{4} \text{He}\)[/tex]

2. [tex]\({ }_{9}^{18} \text{F} \rightarrow { }_{8}^{18} \text{O} + { }_{1}^{0} \text{e}\)[/tex]

3. [tex]\({ }_{6}^{14} \text{C} \rightarrow { }_{7}^{14} \text{N} + { }_{-1}^{0} \text{e}\)[/tex]

4. [tex]\({ }_{66}^{152} \text{Dy} \rightarrow { }_{66}^{152} \text{Dy} + y\)[/tex]

Let's analyze each equation step-by-step to determine which one represents alpha decay:

1. [tex]\({ }_{95}^{241} \text{Am} \rightarrow { }_{93}^{237} \text{Np} + { }_{2}^{4} \text{He}\)[/tex]:
- The element Americium-241 ([tex]\( {}_{95}^{241} \text{Am} \)[/tex]) decays to Neptunium-237 ([tex]\( {}_{93}^{237} \text{Np} \)[/tex]) and an alpha particle ([tex]\( {}_{2}^{4} \text{He} \)[/tex]).
- The atomic number decreases by 2 (from 95 to 93) and the mass number decreases by 4 (from 241 to 237).
- This correctly represents alpha decay.

2. [tex]\({ }_{9}^{18} \text{F} \rightarrow { }_{8}^{18} \text{O} + { }_{1}^{0} \text{e}\)[/tex]:
- The element Fluorine-18 ([tex]\( {}_{9}^{18} \text{F} \)[/tex]) decays to Oxygen-18 ([tex]\( {}_{8}^{18} \text{O} \)[/tex]) and a positron ([tex]\( {}_{1}^{0} \text{e} \)[/tex]).
- This represents beta-plus decay, not alpha decay.

3. [tex]\({ }_{6}^{14} \text{C} \rightarrow { }_{7}^{14} \text{N} + { }_{-1}^{0} \text{e}\)[/tex]:
- The element Carbon-14 ([tex]\( {}_{6}^{14} \text{C} \)[/tex]) decays to Nitrogen-14 ([tex]\( {}_{7}^{14} \text{N} \)[/tex]) and an electron ([tex]\( {}_{-1}^{0} \text{e} \)[/tex]).
- This represents beta-minus decay, not alpha decay.

4. [tex]\({ }_{66}^{152} \text{Dy} \rightarrow { }_{66}^{152} \text{Dy} + y\)[/tex]:
- Dysprosium-152 ([tex]\( {}_{66}^{152} \text{Dy} \)[/tex]) supposedly decays to itself ([tex]\( {}_{66}^{152} \text{Dy} \)[/tex]) and some unspecified particle [tex]\(y\)[/tex].
- This does not represent any standard decay process.

Therefore, the correct equation that represents alpha decay is:

[tex]\({ }_{95}^{241} \text{Am} \rightarrow { }_{93}^{237} \text{Np} + { }_{2}^{4} \text{He}\)[/tex]