which electron transition would result in the emission of light with the longest wavelength, \n\nn=5 -->…

Explanation:

Step1: Recall energy - wavelength relationship

The energy of a photon emitted during an electron - transition is given by $E = h\nu=\frac{hc}{\lambda}$, where $h$ is Planck's constant, $\nu$ is the frequency, $c$ is the speed of light, and $\lambda$ is the wavelength. A lower - energy photon corresponds to a longer wavelength. The energy of an electron transition is $\Delta E=E_f - E_i=- 13.6\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$ eV.

Step2: Calculate energy for each transition

For $n = 5\rightarrow n = 3$: $\Delta E_1=-13.6\left(\frac{1}{3^2}-\frac{1}{5^2}\right)=-13.6\left(\frac{1}{9}-\frac{1}{25}\right)=-13.6\times\frac{25 - 9}{225}=-13.6\times\frac{16}{225}\approx - 0.97$ eV. For $n = 6\rightarrow n = 2$: $\Delta E_2=-13.6\left(\frac{1}{2^2}-\frac{1}{6^2}\right)=-13.6\left(\frac{1}{4}-\frac{1}{36}\right)=-13.6\times\frac{9 - 1}{36}=-13.6\times\frac{8}{36}\approx - 3.02$ eV. For $n = 5\rightarrow n = 2$: $\Delta E_3=-13.6\left(\frac{1}{2^2}-\frac{1}{5^2}\right)=-13.6\left(\frac{1}{4}-\frac{1}{25}\right)=-13.6\times\frac{25 - 4}{100}=-13.6\times\frac{21}{100}=- 2.86$ eV. For $n = 6\rightarrow n = 5$: $\Delta E_4=-13.6\left(\frac{1}{5^2}-\frac{1}{6^2}\right)=-13.6\left(\frac{1}{25}-\frac{1}{36}\right)=-13.6\times\frac{36 - 25}{900}=-13.6\times\frac{11}{900}\approx - 0.16$ eV.

Step3: Identify the transition with lowest energy

The transition with the lowest energy change will emit light with the longest wavelength. Since $\Delta E_4$ (for $n = 6\rightarrow n = 5$) is the smallest in magnitude among the four calculated energy - changes.

Answer:

$n = 6\rightarrow n = 5$