To determine the identity of the resulting atom after the radioactive decay, let's analyze the given decay equation:
[tex]\[
{}_{83}^{209} Bi \rightarrow \_ + {}_2^4 He
\][/tex]
Here, Bismuth-209 ([tex]\( {}_{83}^{209} Bi \)[/tex]) undergoes alpha decay, which involves the emission of an alpha particle ([tex]\( {}_2^4 He \)[/tex]).
An alpha particle consists of 2 protons and 2 neutrons, hence it has:
- A mass number of 4 (2 protons + 2 neutrons)
- An atomic number of 2 (2 protons)
To find the identity of the resulting atom, we need to apply the conservation of mass number and atomic number.
1. Calculating the resulting atom's mass number:
- The initial mass number of Bismuth-209 is 209.
- An alpha particle with a mass number of 4 is emitted.
- Therefore, the mass number of the resulting atom is:
[tex]\[
209 - 4 = 205
\][/tex]
2. Calculating the resulting atom's atomic number:
- The initial atomic number of Bismuth-209 is 83.
- An alpha particle with an atomic number of 2 is emitted.
- Therefore, the atomic number of the resulting atom is:
[tex]\[
83 - 2 = 81
\][/tex]
With the resulting atom having:
- Mass number of 205
- Atomic number of 81
By referring to the periodic table, we find that the element with atomic number 81 is Thallium (Tl).
Thus, the identity of the resulting atom is Thallium-205 ([tex]\( {}_{81}^{205} Tl \)[/tex]).
Therefore, the correct answer is:
A. [tex]\( {}_{81}^{205} Tl \)[/tex]
