To determine which equation represents a balanced alpha emission nuclear equation, let's review the concept of alpha decay. In alpha decay, a nucleus emits an alpha particle, which consists of 2 protons and 2 neutrons (equivalent to a helium nucleus, [tex]\({ }_2^4 He\)[/tex]). This means the mass number of the original nucleus decreases by 4, and the atomic number decreases by 2.
Now, let's analyze each equation:
1. [tex]\({ }_{26}^{52} Fe \longrightarrow{ }_{25}^{52} Mn +{ }_{+1}^0 e\)[/tex]
- The mass number remains 52, and the atomic number decreases from 26 to 25 and releases a positron ([tex]\({ }_{+1}^0 e\)[/tex]).
- This is not an alpha decay. No change in the mass number.
2. [tex]\({ }_{92}^{235} U \longrightarrow{ }_{94}^{239} Pu +{ }_2^4 He\)[/tex]
- The mass number increases from 235 to 239, and the atomic number increases from 92 to 94.
- This is incorrect for alpha decay, as the mass number should decrease by 4, not increase.
3. [tex]\({ }_{83}^{189} Bi \longrightarrow{ }_{81}^{185} Tl +{ }_2^4 He\)[/tex]
- The mass number decreases from 189 to 185 (189-4=185).
- The atomic number decreases from 83 to 81 (83-2=81).
- This matches the criteria for alpha decay.
4. [tex]\({ }_6^{14} C \longrightarrow{ }_7^{14} N +{ }_{-1}^0 e\)[/tex]
- The mass number remains 14, and the atomic number increases from 6 to 7 and releases a beta particle ([tex]\({ }_{-1}^0 e\)[/tex]).
- This is not an alpha decay. No change in the mass number.
Thus, the correct equation representing a balanced alpha emission nuclear equation is:
[tex]\({ }_{83}^{189} Bi \longrightarrow{ }_{81}^{185} Tl +{ }_2^4 He\)[/tex]
So, the correct answer is [tex]\( \boxed{3} \)[/tex].
