To determine which option represents a beta decay, let's closely examine the provided choices and understand what constitutes a beta decay.
Beta Decay:
- Beta decay involves the transformation of a neutron into a proton, which increases the atomic number by 1 while the mass number remains unchanged. During this process, a beta particle (electron) is emitted.
- The general form of beta decay can be written as:
[tex]\[
{}_{Z}^{A}X \rightarrow {}_{Z+1}^{A}Y + {}_{-1}^{0}e
\][/tex]
Let's now investigate each option:
Option A:
[tex]\[ {}_{83}^{212} Bi \rightarrow {}_{84}^{212} Po + {}_{-1}^{0} e \][/tex]
- The atomic number of Bi is 83, which increases to 84 in Po.
- The mass number remains the same (212).
- Emits a beta particle ([tex]${ }_{-1}^{0} e$[/tex]).
This matches the criteria for a beta decay. Hence, Option A represents a beta decay.
Option B:
[tex]\[ {}_{28}^{60} Ni \rightarrow {}_{28}^{60} Ni + y \][/tex]
- The atomic number of Ni remains unchanged (28).
- The mass number remains unchanged (60).
- Emits gamma radiation (y), which indicates a gamma decay, not a beta decay.
This does not match the criteria for a beta decay.
Option C:
[tex]\[ {}_{86}^{220} Rn \rightarrow {}_{84}^{216} Po + {}_{2}^{4} He \][/tex]
- The atomic number of Rn is 86, which decreases to 84 in Po.
- The mass number decreases from 220 to 216.
- Emits an alpha particle ([tex]${ }_{2}^{4} He$[/tex]).
This indicates an alpha decay, not a beta decay.
Option D:
[tex]\[ {}_{93}^{235} Np \rightarrow {}_{91}^{231} Pa + \alpha \][/tex]
- The atomic number of Np is 93, which decreases to 91 in Pa.
- The mass number decreases from 235 to 231.
- Emits an alpha particle ([tex]$\alpha$[/tex]).
This is another example of an alpha decay, not a beta decay.
Based on the examination of all options, we can conclude that Option A correctly represents a beta decay. Therefore, the correct answer is:
[tex]\[ \boxed{1} \][/tex]
