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\n| |n = 6→n = 2|n = 6→n = 1|\n|--|--|--|\n|a. photon energy|(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})|(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})|\n|b. frequency|\n|c. wavelength|\n

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\n| |n = 6→n = 2|n = 6→n = 1|\n|--|--|--|\n|a. photon energy|(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})|(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})|\n|b. frequency|\n|c. wavelength|\n

Answer

Explanation:

Step1: Recall the Rydberg - energy formula

The energy of a photon emitted during an electron transition in a hydrogen atom is given by $E = 13.6\ eV\times\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)$, where $n_{i}$ is the initial energy - level and $n_{f}$ is the final energy - level.

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

$E_{1}=13.6\ eV\times\left(\frac{1}{2^{2}}-\frac{1}{6^{2}}\right)=13.6\ eV\times\left(\frac{1}{4}-\frac{1}{36}\right)=13.6\ eV\times\frac{9 - 1}{36}=13.6\ eV\times\frac{8}{36}\approx3.02\ eV$

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

$E_{2}=13.6\ eV\times\left(\frac{1}{1^{2}}-\frac{1}{6^{2}}\right)=13.6\ eV\times\left(1-\frac{1}{36}\right)=13.6\ eV\times\frac{36 - 1}{36}=13.6\ eV\times\frac{35}{36}\approx13.1\ eV$

Step2: Use the energy - frequency relation

The energy of a photon is also given by $E = h\nu$, where $h = 4.136\times10^{-15}\ eV\cdot s$ is Planck's constant and $\nu$ is the frequency.

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

$\nu_{1}=\frac{E_{1}}{h}=\frac{3.02\ eV}{4.136\times10^{-15}\ eV\cdot s}\approx7.3\times10^{14}\ Hz$

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

$\nu_{2}=\frac{E_{2}}{h}=\frac{13.1\ eV}{4.136\times10^{-15}\ eV\cdot s}\approx3.17\times10^{15}\ Hz$

Step3: Use the frequency - wavelength relation

The speed of light $c = 3\times10^{8}\ m/s$, and $c=\lambda\nu$, so $\lambda=\frac{c}{\nu}$.

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

$\lambda_{1}=\frac{c}{\nu_{1}}=\frac{3\times10^{8}\ m/s}{7.3\times10^{14}\ Hz}\approx411\ nm$

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

$\lambda_{2}=\frac{c}{\nu_{2}}=\frac{3\times10^{8}\ m/s}{3.17\times10^{15}\ Hz}\approx94.6\ nm$

Answer:

Transition Photon Energy (eV) Frequency (Hz) Wavelength (nm)
$n = 6\rightarrow n = 2$ $\approx3.02$ $\approx7.3\times10^{14}$ $\approx411$
$n = 6\rightarrow n = 1$ $\approx13.1$ $\approx3.17\times10^{15}$ $\approx94.6$