To determine which electronic transition in a hydrogen atom would result in the emission of a photon with the longest wavelength, we'll analyze each of the given transitions from the initial state [tex]\( n_i = 3 \)[/tex] to the final state [tex]\( n_f \)[/tex].
For each transition, we calculate the wavelength of the emitted photon. The transitions provided are:
- [tex]\( n=3 \rightarrow n=1 \)[/tex]
- [tex]\( n=3 \rightarrow n=4 \)[/tex]
- [tex]\( n=3 \rightarrow n=5 \)[/tex]
- [tex]\( n=3 \rightarrow n=2 \)[/tex]
We will compare the wavelengths for these transitions, as the longest wavelength corresponds to the smallest energy change.
Here are the results:
1. For the transition [tex]\( n=3 \rightarrow n=1 \)[/tex]:
- Wavelength: [tex]\( 1.0251754318726755 \times 10^{-7} \)[/tex] meters
2. For the transition [tex]\( n=3 \rightarrow n=4 \)[/tex]:
- Wavelength: [tex]\( -1.8746065039957495 \times 10^{-6} \)[/tex] meters
3. For the transition [tex]\( n=3 \rightarrow n=5 \)[/tex]:
- Wavelength: [tex]\( -1.2814692898408445 \times 10^{-6} \)[/tex] meters
4. For the transition [tex]\( n=3 \rightarrow n=2 \)[/tex]:
- Wavelength: [tex]\( 6.561122763985122 \times 10^{-7} \)[/tex] meters
Among these values, the transition [tex]\( n=3 \rightarrow n=4 \)[/tex] has the longest wavelength.
Therefore, the electronic transition in a hydrogen atom that would result in the emission of a photon with the longest wavelength is:
[tex]\[ \boxed{n=3 \rightarrow n=4} \][/tex]
