2. compare and contrast the photons of light produced when an electron in a hydrogen atom makes the…

Explanation:

Step 1: Recall the Rydberg - energy formula

The energy of a photon emitted during an electron transition in a hydrogen atom is given by $E = h\nu= - 13.6\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)\text{ eV}$, where $n_i$ is the initial energy - level and $n_f$ is the final energy - level, $h = 6.626\times10^{-34}\text{ J}\cdot\text{s}$ is Planck's constant, and $\nu$ is the frequency of the photon.

Step 2: Calculate photon energy for $n = 6\rightarrow n = 2$

For $n_i = 6$ and $n_f = 2$: [ \begin{align*} E_{6\rightarrow2}&=- 13.6\left(\frac{1}{2^2}-\frac{1}{6^2}\right)\text{ eV}\ &=-13.6\left(\frac{1}{4}-\frac{1}{36}\right)\text{ eV}\ &=-13.6\times\frac{9 - 1}{36}\text{ eV}\ &=-13.6\times\frac{8}{36}\text{ eV}\ &\approx - 3.02\text{ eV} \end{align*} ] The magnitude of the energy of the photon is $|E_{6\rightarrow2}|=3.02\text{ eV}=3.02\times1.6\times10^{-19}\text{ J}=4.83\times10^{-19}\text{ J}$

Step 3: Calculate photon energy for $n = 6\rightarrow n = 1$

For $n_i = 6$ and $n_f = 1$: [ \begin{align*} E_{6\rightarrow1}&=-13.6\left(\frac{1}{1^2}-\frac{1}{6^2}\right)\text{ eV}\ &=-13.6\left(1-\frac{1}{36}\right)\text{ eV}\ &=-13.6\times\frac{36 - 1}{36}\text{ eV}\ &=-13.6\times\frac{35}{36}\text{ eV}\ &\approx - 13.2\text{ eV} \end{align*} ] The magnitude of the energy of the photon is $|E_{6\rightarrow1}| = 13.2\text{ eV}=13.2\times1.6\times10^{-19}\text{ J}=2.11\times10^{-18}\text{ J}$

Step 4: Calculate frequency using $E = h\nu$

For $n = 6\rightarrow n = 2$: [ \nu_{6\rightarrow2}=\frac{E_{6\rightarrow2}}{h}=\frac{4.83\times10^{-19}\text{ J}}{6.626\times10^{-34}\text{ J}\cdot\text{s}}\approx7.3\times10^{14}\text{ Hz} ] For $n = 6\rightarrow n = 1$: [ \nu_{6\rightarrow1}=\frac{E_{6\rightarrow1}}{h}=\frac{2.11\times10^{-18}\text{ J}}{6.626\times10^{-34}\text{ J}\cdot\text{s}}\approx3.18\times10^{15}\text{ Hz} ]

Step 5: Calculate wavelength using $\lambda=\frac{c}{\nu}$ where $c = 3\times10^{8}\text{ m/s}$

For $n = 6\rightarrow n = 2$: [ \lambda_{6\rightarrow2}=\frac{c}{\nu_{6\rightarrow2}}=\frac{3\times10^{8}\text{ m/s}}{7.3\times10^{14}\text{ Hz}}\approx4.11\times10^{-7}\text{ m} = 411\text{ nm} ] For $n = 6\rightarrow n = 1$: [ \lambda_{6\rightarrow1}=\frac{c}{\nu_{6\rightarrow1}}=\frac{3\times10^{8}\text{ m/s}}{3.18\times10^{15}\text{ Hz}}\approx9.43\times10^{-8}\text{ m}=94.3\text{ nm} ]

Answer:

a. Photon energy: $n = 6\rightarrow n = 2$: $3.02\text{ eV}$ or $4.83\times10^{-19}\text{ J}$; $n = 6\rightarrow n = 1$: $13.2\text{ eV}$ or $2.11\times10^{-18}\text{ J}$ b. Frequency: $n = 6\rightarrow n = 2$: $7.3\times10^{14}\text{ Hz}$; $n = 6\rightarrow n = 1$: $3.18\times10^{15}\text{ Hz}$ c. Wavelength: $n = 6\rightarrow n = 2$: $411\text{ nm}$; $n = 6\rightarrow n = 1$: $94.3\text{ nm}$