Which of the following represents an alpha decay?

A. [tex]\({ }_{93}^{235} \text{Np} \rightarrow { }_{91}^{231} \text{Pa} + \alpha\)[/tex]

B. [tex]\({ }_{83}^{212} \text{Bi} \rightarrow { }_{84}^{212} \text{Po} + { }_{-1}^{0} \text{e}\)[/tex]

C. [tex]\({ }_{93}^{239} \text{Np} \rightarrow { }_{94}^{239} \text{Pu} + \beta\)[/tex]

D. [tex]\({ }_{28}^{60} \text{Ni} \rightarrow { }_{28}^{60} \text{Ni} + \gamma\)[/tex]



Answer :

To determine which option represents an alpha decay, it is important to understand the characteristics of an alpha decay. An alpha decay involves the emission of an alpha particle, which is denoted as [tex]\({}^4_2 He\)[/tex]. This causes a reduction in both the atomic number and the mass number of the original nucleus. The atomic number decreases by 2 and the mass number decreases by 4.

Let's analyze each option in detail:

A. [tex]\({}_{93}^{235} Np \rightarrow {}_{91}^{231} Pa + \alpha\)[/tex]
- Here, the atomic number of the original nucleus [tex]\(Np\)[/tex] is 93, and it decreases to 91 in the product [tex]\(Pa\)[/tex] (decrease by 2).
- The mass number of the original nucleus [tex]\(Np\)[/tex] is 235, and it decreases to 231 in the product [tex]\(Pa\)[/tex] (decrease by 4).
- This change corresponds to the emissions of an alpha particle ([tex]\({}_2^4 He\)[/tex]).

B. [tex]\({}_{83}^{212} Bi \rightarrow {}_{84}^{212} Po + {}_{-1}^0 e\)[/tex]
- Here, the atomic number of the original nucleus [tex]\(Bi\)[/tex] is 83, and it increases to 84 in the product [tex]\(Po\)[/tex] (increase by 1).
- The mass number remains unchanged at 212.
- This change corresponds to beta decay, not alpha decay.

C. [tex]\({}_{93}^{239} Np \rightarrow {}_{94}^{239} Pu + \beta\)[/tex]
- Here, the atomic number of the original nucleus [tex]\(Np\)[/tex] is 93, and it increases to 94 in the product [tex]\(Pu\)[/tex] (increase by 1).
- The mass number remains unchanged at 239.
- This change also corresponds to beta decay, not alpha decay.

D. [tex]\({}_{28}^{60} Ni \rightarrow {}_{28}^{60} Ni + \gamma\)[/tex]
- Here, the atomic number remains 28, and the mass number remains 60.
- This nuclei emits a gamma-ray, but there is no change in the atomic number or mass number, corresponding to a gamma decay, not an alpha decay.

From the analysis above, it is clear that:

Option A: [tex]\({}_{93}^{235} Np \rightarrow {}_{91}^{231} Pa + \alpha\)[/tex] is the correct representation of an alpha decay.

Therefore, the correct answer is:
[tex]\[ \boxed{1} \][/tex]