Answer :
To determine which equation represents alpha decay, let's review the concept of alpha decay. Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle, consisting of 2 protons and 2 neutrons, resulting in the formation of a new nucleus with both the atomic number (number of protons) and mass number (total number of protons and neutrons) lower by 2 and 4, respectively.
Let's analyze each given option to identify the correct alpha decay process.
Option A: [tex]\( \text{ }^{222}_{86} \mathrm{Rn} \rightarrow \mathrm{ }^{218}_{84} \mathrm{Po} + X \)[/tex]
- The initial atom is radon-222 with 86 protons.
- The resulting atom is polonium-218 with 84 protons.
- The change in atomic number is [tex]\( 86 - 84 = 2 \)[/tex], indicating the loss of 2 protons, which is consistent with an alpha particle.
- The change in mass number is [tex]\( 222 - 218 = 4 \)[/tex], indicating the loss of an alpha particle with 2 protons and 2 neutrons.
- Thus, [tex]\( X \)[/tex] represents an alpha particle: [tex]\( \mathrm{ }^{4}_{2} \mathrm{He} \)[/tex].
Option A represents the correct alpha decay as it satisfies the proper emission of an alpha particle.
Option B: [tex]\( \text{ }^{116}_{49} \mathrm{In} \rightarrow \mathrm{ }^{116}_{50} \mathrm{Sn} + X \)[/tex]
- The initial atom is indium-116 with 49 protons.
- The resulting atom is tin-116 with 50 protons.
- The change in atomic number is [tex]\( 50 - 49 = 1 \)[/tex], indicating the gain of 1 proton, not consistent with alpha decay.
- Additionally, the mass number remains unchanged, which is not consistent with alpha decay.
Option B does not represent alpha decay.
Option C: [tex]\( \text{ }^{38}_{19} \mathrm{K} \rightarrow \mathrm{ }^{38}_{18} \mathrm{Ar} + X \)[/tex]
- The initial atom is potassium-38 with 19 protons.
- The resulting atom is argon-38 with 18 protons.
- The change in atomic number is [tex]\( 19 - 18 = 1 \)[/tex], indicating the loss of 1 proton, and the mass number is unchanged, which is not consistent with alpha decay.
Option C does not represent alpha decay.
Option D: [tex]\( \text{ }^{334}_{91} \mathrm{Th} \rightarrow \mathrm{ }^{234}_{90} \mathrm{Pa} + X \)[/tex]
- The initial atom is thorium-334 with 91 protons.
- The resulting atom is protactinium-234 with 90 protons.
- The change in atomic number is [tex]\( 91 - 90 = 1 \)[/tex], and the mass number changes significantly from 334 to 234, which is not consistent with alpha decay.
Option D does not represent alpha decay.
Therefore, the correct equation representing alpha decay is:
A) [tex]\( \text{ }^{222}_{86} \mathrm{Rn} \rightarrow \mathrm{ }^{218}_{84} \mathrm{Po} + \mathrm{ }^{4}_{2} \mathrm{He} \)[/tex]
Let's analyze each given option to identify the correct alpha decay process.
Option A: [tex]\( \text{ }^{222}_{86} \mathrm{Rn} \rightarrow \mathrm{ }^{218}_{84} \mathrm{Po} + X \)[/tex]
- The initial atom is radon-222 with 86 protons.
- The resulting atom is polonium-218 with 84 protons.
- The change in atomic number is [tex]\( 86 - 84 = 2 \)[/tex], indicating the loss of 2 protons, which is consistent with an alpha particle.
- The change in mass number is [tex]\( 222 - 218 = 4 \)[/tex], indicating the loss of an alpha particle with 2 protons and 2 neutrons.
- Thus, [tex]\( X \)[/tex] represents an alpha particle: [tex]\( \mathrm{ }^{4}_{2} \mathrm{He} \)[/tex].
Option A represents the correct alpha decay as it satisfies the proper emission of an alpha particle.
Option B: [tex]\( \text{ }^{116}_{49} \mathrm{In} \rightarrow \mathrm{ }^{116}_{50} \mathrm{Sn} + X \)[/tex]
- The initial atom is indium-116 with 49 protons.
- The resulting atom is tin-116 with 50 protons.
- The change in atomic number is [tex]\( 50 - 49 = 1 \)[/tex], indicating the gain of 1 proton, not consistent with alpha decay.
- Additionally, the mass number remains unchanged, which is not consistent with alpha decay.
Option B does not represent alpha decay.
Option C: [tex]\( \text{ }^{38}_{19} \mathrm{K} \rightarrow \mathrm{ }^{38}_{18} \mathrm{Ar} + X \)[/tex]
- The initial atom is potassium-38 with 19 protons.
- The resulting atom is argon-38 with 18 protons.
- The change in atomic number is [tex]\( 19 - 18 = 1 \)[/tex], indicating the loss of 1 proton, and the mass number is unchanged, which is not consistent with alpha decay.
Option C does not represent alpha decay.
Option D: [tex]\( \text{ }^{334}_{91} \mathrm{Th} \rightarrow \mathrm{ }^{234}_{90} \mathrm{Pa} + X \)[/tex]
- The initial atom is thorium-334 with 91 protons.
- The resulting atom is protactinium-234 with 90 protons.
- The change in atomic number is [tex]\( 91 - 90 = 1 \)[/tex], and the mass number changes significantly from 334 to 234, which is not consistent with alpha decay.
Option D does not represent alpha decay.
Therefore, the correct equation representing alpha decay is:
A) [tex]\( \text{ }^{222}_{86} \mathrm{Rn} \rightarrow \mathrm{ }^{218}_{84} \mathrm{Po} + \mathrm{ }^{4}_{2} \mathrm{He} \)[/tex]
