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

2. compare and contrast the photons of light produced when an electron in a hydrogen atom makes the transition 6 → 2 vs. 6 → 1 by calculating the following quantities. show your work clearly for each calculation. the introductory pages of this lab (and section 3.2 of your textbook) have examples of these calculations.\n\nn = 6→n = 2\ta. photon energy\n2.18×10^(-18)(1/2² - 1/6²)\n-2.18×10^(-18)(1/4 - 1/36)\n2.15×10^(-19)\ne=-4.84×10^(-19) j\nb. frequency\nν=\n6.63×10^(-34)\n\nc. wavelength\n=3×10^8/λ\nλ=\n\nn = 6→n = 1\n-2.18×10^(-18)(1/1² - 1/6²)\n3×10^8/λ\nλ=\n\nd. which of these transitions (6 → 2 or 6 → 1) produces light that is (just barely) within the visible spectrum? in what part of the spectrum is the light from the other transition?\n6→2 produces visible light in the spectrum (6→1 produces infrared or ultraviolet light which cannot be seen (lyman series)\ne. consider an electron making the transition 7 → 2. without doing specific calculations, predict how the wavelength of the light from the 7 → 2 transition would compare to the wavelength for the 6 → 2 transition. in what part of the spectrum would the light be for the 7 → 2 transition?\n7→2 would be part of the balmer series

2. compare and contrast the photons of light produced when an electron in a hydrogen atom makes the transition 6 → 2 vs. 6 → 1 by calculating the following quantities. show your work clearly for each calculation. the introductory pages of this lab (and section 3.2 of your textbook) have examples of these calculations.\n\nn = 6→n = 2\ta. photon energy\n2.18×10^(-18)(1/2² - 1/6²)\n-2.18×10^(-18)(1/4 - 1/36)\n2.15×10^(-19)\ne=-4.84×10^(-19) j\nb. frequency\nν=\n6.63×10^(-34)\n\nc. wavelength\n=3×10^8/λ\nλ=\n\nn = 6→n = 1\n-2.18×10^(-18)(1/1² - 1/6²)\n3×10^8/λ\nλ=\n\nd. which of these transitions (6 → 2 or 6 → 1) produces light that is (just barely) within the visible spectrum? in what part of the spectrum is the light from the other transition?\n6→2 produces visible light in the spectrum (6→1 produces infrared or ultraviolet light which cannot be seen (lyman series)\ne. consider an electron making the transition 7 → 2. without doing specific calculations, predict how the wavelength of the light from the 7 → 2 transition would compare to the wavelength for the 6 → 2 transition. in what part of the spectrum would the light be for the 7 → 2 transition?\n7→2 would be part of the balmer series

Answer

Explanation:

Step1: Calculate photon energy for $n = 6\rightarrow n = 2$

The formula for photon energy in a hydrogen - atom electron transition is $E=- 2.18\times10^{-18}\left(\frac{1}{n_f^{2}}-\frac{1}{n_i^{2}}\right)$. Here, $n_i = 6$ and $n_f = 2$. So, $E=-2.18\times10^{-18}\left(\frac{1}{2^{2}}-\frac{1}{6^{2}}\right)=-2.18\times10^{-18}\left(\frac{1}{4}-\frac{1}{36}\right)=-2.18\times10^{-18}\left(\frac{9 - 1}{36}\right)=-2.18\times10^{-18}\times\frac{8}{36}\approx - 4.84\times10^{-19}\text{ J}$.

Step2: Calculate photon energy for $n = 6\rightarrow n = 1$

Using the same formula with $n_i = 6$ and $n_f = 1$, we have $E=-2.18\times10^{-18}\left(\frac{1}{1^{2}}-\frac{1}{6^{2}}\right)=-2.18\times10^{-18}\left(1-\frac{1}{36}\right)=-2.18\times10^{-18}\times\frac{35}{36}\approx - 2.12\times10^{-18}\text{ J}$.

Step3: Calculate frequency for $n = 6\rightarrow n = 2$

We know that $E = h\nu$, so $\nu=\frac{E}{h}$. Given $E=-4.84\times10^{-19}\text{ J}$ and $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$, $\nu=\frac{4.84\times10^{-19}}{6.63\times10^{-34}}\approx7.30\times10^{14}\text{ Hz}$.

Step4: Calculate frequency for $n = 6\rightarrow n = 1$

Using $E=-2.12\times10^{-18}\text{ J}$ and $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$, $\nu=\frac{2.12\times10^{-18}}{6.63\times10^{-34}}\approx3.19\times10^{15}\text{ Hz}$.

Step5: Calculate wavelength for $n = 6\rightarrow n = 2$

Since $c=\lambda\nu$, $\lambda=\frac{c}{\nu}$. With $c = 3\times10^{8}\text{ m/s}$ and $\nu = 7.30\times10^{14}\text{ Hz}$, $\lambda=\frac{3\times10^{8}}{7.30\times10^{14}}\approx4.11\times10^{-7}\text{ m}=411\text{ nm}$.

Step6: Calculate wavelength for $n = 6\rightarrow n = 1$

Using $c = 3\times10^{8}\text{ m/s}$ and $\nu = 3.19\times10^{15}\text{ Hz}$, $\lambda=\frac{3\times10^{8}}{3.19\times10^{15}}\approx9.40\times10^{-8}\text{ m}=94.0\text{ nm}$.

Step7: Determine the spectrum for transitions

The visible - light spectrum ranges from approximately $400 - 700\text{ nm}$. The transition $n = 6\rightarrow n = 2$ with $\lambda\approx411\text{ nm}$ is in the visible spectrum. The transition $n = 6\rightarrow n = 1$ with $\lambda\approx94.0\text{ nm}$ is in the ultraviolet (UV) spectrum.

Step8: Predict for $n = 7\rightarrow n = 2$

The energy of a transition is given by $E=-2.18\times10^{-18}\left(\frac{1}{n_f^{2}}-\frac{1}{n_i^{2}}\right)$. The transition $n = 7\rightarrow n = 2$ has a larger $n_i$ than $n = 6\rightarrow n = 2$. A larger $n_i$ for the same $n_f$ means a smaller energy difference. Since $E = h\nu=\frac{hc}{\lambda}$, a smaller energy difference corresponds to a longer wavelength. The $n = 7\rightarrow n = 2$ transition is also part of the Balmer series (transitions to $n = 2$) and is in the visible spectrum.

Answer:

For $n = 6\rightarrow n = 2$:

  • Photon energy: $-4.84\times10^{-19}\text{ J}$
  • Frequency: $7.30\times10^{14}\text{ Hz}$
  • Wavelength: $411\text{ nm}$
  • Spectrum: Visible

For $n = 6\rightarrow n = 1$:

  • Photon energy: $-2.12\times10^{-18}\text{ J}$
  • Frequency: $3.19\times10^{15}\text{ Hz}$
  • Wavelength: $94.0\text{ nm}$
  • Spectrum: Ultraviolet

For $n = 7\rightarrow n = 2$:

  • Wavelength is longer than that of $n = 6\rightarrow n = 2$
  • Spectrum: Visible (Balmer series)