To determine the correct equation representing the radioactive decay of [tex]\( \,^{226}_{88}\text{Ra} \)[/tex], we need to understand the different types of decay processes:
1. Alpha Decay: This involves the emission of an alpha particle ([tex]\( \,_{2}^4\text{He} \)[/tex]). An alpha particle has 2 protons and 2 neutrons, so the atomic number decreases by 2 and the mass number decreases by 4.
2. Beta Decay (Beta-minus): This involves the transformation of a neutron into a proton with the emission of a beta particle ([tex]\( \,_{-1}^0\text{e} \)[/tex]). The atomic number increases by 1 while the mass number remains unchanged.
3. Beta Decay (Beta-plus or positron emission): This involves the transformation of a proton into a neutron with the emission of a positron ([tex]\( \,_{+1}^0\text{e} \)[/tex]). The atomic number decreases by 1 while the mass number remains unchanged.
4. Neutron Emission: This involves the emission of a neutron ([tex]\( \,_{0}^1\text{n} \)[/tex]). The atomic number remains unchanged while the mass number decreases by 1.
Now, let's analyze each option with the given element [tex]\( \,^{226}_{88}\text{Ra} \)[/tex]:
- Option A: [tex]\( \,^{226}_{88}\text{Ra} \rightarrow \,^{222}_{86}\text{Rn} + \,_{2}^4\text{He} \)[/tex]
- This represents alpha decay. The atomic number decreases from 88 to 86 (88 - 2) and the mass number decreases from 226 to 222 (226 - 4). This is a correct alpha decay equation.
- Option B: [tex]\( \,^{226}_{88}\text{Ra} \rightarrow \,^{226}_{89}\text{Ac} + \,_{-1}^0\text{e} \)[/tex]
- This represents beta-minus decay. The atomic number increases from 88 to 89 (due to the emission of a beta particle), but the mass number remains the same. This is not possible for [tex]\( \,^{226}_{88}\text{Ra} \)[/tex].
- Option C: [tex]\( \,^{226}_{88}\text{Ra} \rightarrow \,^{226}_{87}\text{Fr} + \,_{+1}^0\text{e} \)[/tex]
- This represents beta-plus decay. The atomic number decreases from 88 to 87 (due to the emission of a positron), but the mass number remains the same. This is not correct for [tex]\( \,^{226}_{88}\text{Ra} \)[/tex].
- Option D: [tex]\( \,^{226}_{88}\text{Ra} \rightarrow \,^{225}_{88}\text{Ra} + \,_{0}^1\text{n} \)[/tex]
- This represents neutron emission. The atomic number remains unchanged but the mass number decreases by 1. This is not a typical decay for [tex]\( \,^{226}_{88}\text{Ra} \)[/tex].
Given the above analysis, the correct decay process for [tex]\( \,^{226}_{88}\text{Ra} \)[/tex] is Alpha Decay, as described in Option A:
[tex]\( \boxed{ \,^{226}_{88}\text{Ra} \rightarrow \,^{222}_{86}\text{Rn} + \,_{2}^4\text{He} } \)[/tex]
