An atomic emission spectrum of hydrogen shows three wavelengths: 121.5

Tara Alvarado 2022-01-12 Answered
An atomic emission spectrum of hydrogen shows three wavelengths: 121.5 nm , 102.6 nm, and 97.23 nm . Assign these wavelengths to transitions in the hydrogen atom.

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Expert Answer

Maricela Alarcon
Answered 2022-01-13 Author has 5409 answers

Step 1
Given the emission spectrum of hydrogen emits the wavelengths (a) 121.5nm, (b) 102.6nm, and (c) 97.23nm.
Determine the energy level transitions of each wavelength.
Solution: Use the wavelength to calculate the \(\displaystyle\triangle{E}\) in the atom. Use \(\displaystyle\triangle_{{{a}to{m}}}\) to determine the \(\displaystyle{n}_{{{i}}}\ {\quad\text{and}\quad}\ {n}_{{{f}}}\).
Step 2
Then, the energy of the photon can be calculated using
\(\displaystyle{E}_{{{p}{h}{o}to{n}}}=-{\frac{{{h}{c}}}{{\lambda}}}\)
where \(\displaystyle{h}={6.626}\times{10}^{{-{34}}}{J}\cdot{s}\ {\quad\text{and}\quad}\ {c}={3.00}\times{10}^{{{8}}}\frac{{m}}{{s}}\). The negative sign indicates that the energy in the atom is released as a photon.
Since energy is conserved, the emitted energy of the photon is equal to the negative change in energy of the atom.
\(\displaystyle\triangle{E}_{{{a}to{m}}}=-{E}_{{{p}{h}{o}to{n}}}\)
Step 3
Note that since \(\displaystyle\lambda{ < }{400}{n}{m}\) is in the ultraviolet region. Ultraviolet emissions correspond to electron transition with a final n at 1. Therefore, \(\displaystyle{n}_{{{f}}}={1}\) of all the transitions. The formula for \(\displaystyle\triangle{E}\) for a hydrogen atom is:
\(\displaystyle\triangle{E}_{{{a}to{m}}}=-{2.18}\times{10}^{{-{18}}}{J}\cdot{\frac{{{1}}}{{{{n}_{{{f}}}^{{{2}}}}}}}-{\left(-{2.18}\times{10}^{{-{18}}}{J}\cdot{\frac{{{1}}}{{{{n}_{{{i}}}^{{{2}}}}}}}\right)}\)
isolating \(\displaystyle{n}_{{{f}}}\)
\(\displaystyle{n}_{{{i}}}=\sqrt{{{\frac{{\triangle{E}_{{{a}to{m}}}+{2.18}\times{10}^{{-{18}}}{J}\cdot{\frac{{{1}}}{{{{n}_{{{f}}}^{{{2}}}}}}}}}{{{2.18}\times{10}^{{-{18}}}{J}}}}}}\)
Step 4
(a) 121.5nm
Calculate E-photon
\(\displaystyle{E}_{{{p}{h}{o}to{n}}}=-{\frac{{{6.626}\times{10}^{{-{34}}}{J}\cdot{s}\cdot{3.00}\times{10}^{{{8}}}\frac{{m}}{{s}}}}{{{121.5}{n}{m}\cdot{\frac{{{1}\times{10}^{{-{9}}}{m}}}{{{1}{n}{m}}}}}}}\)
Calculate E-atom
\(\displaystyle{E}_{{{a}to{m}}}=-{E}_{{{p}{h}{o}to{n}}}={1.61605}\times{10}^{{-{18}}}{J}\)
Calculate for \(\displaystyle{n}_{{{i}}}\) using \(\displaystyle{n}_{{{f}}}={3}\)
\(\displaystyle{n}_{{{i}}}=\sqrt{{{\frac{{{2.18}\times{10}^{{-{18}}}{J}}}{{{1.61605}\times{10}^{{-{18}}}{J}+{2.18}\times{10}^{{-{18}}}{J}\cdot{\frac{{{1}}}{{{1}^{{{2}}}}}}}}}}}={1.96611}\)
\(\displaystyle{n}={2}\Rightarrow{n}={1}\)
Step 5
(b) 102.6nm
Calculate E-photon
\(\displaystyle{E}_{{{p}{h}{o}to{n}}}=-{\frac{{{6.626}\times{10}^{{-{34}}}{J}\cdot{s}\cdot{3.00}\times{10}^{{{8}}}\frac{{m}}{{s}}}}{{{102.6}{n}{m}\cdot{\frac{{{1}\times{10}^{{-{9}}}{m}}}{{{1}{n}{m}}}}}}}=-{1.937427}\times{10}^{{-{18}}}{J}\)
Calculate E-atom
\(\displaystyle{E}_{{{a}to{m}}}=-{E}_{{{p}{h}{o}to{n}}}={1.937427}\times{10}^{{-{18}}}{J}\)
Calculate for \(\displaystyle{n}_{{{i}}}\) using \(\displaystyle{n}_{{{f}}}={3}\)
\(\displaystyle{n}_{{{i}}}=\sqrt{{{\frac{{{2.18}\times{10}^{{-{18}}}{J}}}{{{1.937427}\times{10}^{{-{18}}}{J}+{2.18}\times{10}^{{-{18}}}{J}\cdot{\frac{{{1}}}{{{1}^{{{2}}}}}}}}}}}={2.9978}\)
\(\displaystyle{n}={3}\Rightarrow{n}={1}\)
Step 6
(c) 97.23nm
Calculate E-photon
\(\displaystyle{E}_{{{p}{h}{o}to{n}}}=-{\frac{{{6.626}\times{10}^{{-{34}}}{J}\cdot{s}\cdot{3.00}\times{10}^{{{8}}}\frac{{m}}{{s}}}}{{{97.23}{n}{m}\cdot{\frac{{{1}\times{10}^{{-{9}}}{m}}}{{{1}{n}{m}}}}}}}=-{2.0444}\times{10}^{{-{18}}}{J}\)

Calculate \(\displaystyle{E}_{{{a}to{m}}}\)
\(\displaystyle{E}_{{{a}to{m}}}=-{E}_{{{p}{h}{o}to{n}}}=-{2.0444}\times{10}^{{-{18}}}{J}\)
Calculate for \(\displaystyle{n}_{{{i}}}\) using \(\displaystyle{n}_{{{f}}}={3}\).
\(\displaystyle{n}_{{{i}}}=\sqrt{{{\frac{{{2.18}\times{10}^{{-{18}}}{J}}}{{-{2.0444}\times{10}^{{-{18}}}{J}+{2.18}\times{10}^{{-{18}}}{J}\cdot{\frac{{{1}}}{{{1}^{{{2}}}}}}}}}}}={4.01002}\)
\(\displaystyle{n}={4}\Rightarrow{n}={1}\)

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temzej9
Answered 2022-01-14 Author has 5601 answers
We have that wavelengths to transitions in the hydrogen atom.
\(\displaystyle{n}{f}={2},{n}{i}={1}\Rightarrow{121.5}{n}{m}\)
\(\displaystyle{n}{i}={1},{n}{f}={3}\Rightarrow{102.6}{n}{m}\)
\(\displaystyle{n}{i}={1},{n}{f}={4}\Rightarrow{97.2}{n}{m}\)
From the question we are told that
121.5 nm,102.6 nm, and 97.23 nm
Generally the equation for wavelength is mathematically given as
\(\displaystyle{\frac{{{1}}}{{\lambda}}}={R}_{{{H}}}\cdot{\left({\frac{{{1}}}{{{n}{f}^{{{2}}}}}}-{\frac{{{1}}}{{{n}{f}^{{{2}}}}}}\right)}\)
\(\displaystyle{\frac{{{1}}}{{\lambda}}}={1.097}\cdot{10}^{{{7}}}\cdot{\frac{{{1}}}{{{n}{f}^{{{2}}}}}}-{\frac{{{1}}}{{{n}{i}^{{{2}}}}}}\)
Therefore
For \(\displaystyle{n}{f}={2},{n}{i}={1}\)
\(\displaystyle{\frac{{{1}}}{{\lambda}}}={1.097}\cdot{10}^{{{7}}}\cdot{\frac{{{1}}}{{{2}^{{{2}}}}}}-{\frac{{{1}}}{{{1}^{{{2}}}}}}\)
\(\displaystyle{\frac{{{1}}}{{\lambda}}}={121.5}{n}{m}\)
For \(\displaystyle{n}{i}={1},{n}{f}={3}\)
\(\displaystyle{\frac{{{1}}}{{\lambda}}}={1.097}\cdot{10}^{{{7}}}\cdot{\frac{{{3}}}{{{1}^{{{2}}}}}}-{\frac{{{1}}}{{{1}^{{{2}}}}}}\)
\(\displaystyle{\frac{{{1}}}{{\lambda}}}={102.6}{n}{m}\)
For \(\displaystyle{n}{i}={1},{n}{f}={4}\)
\(\displaystyle{\frac{{{1}}}{{\lambda}}}={1.097}\cdot{10}^{{{7}}}\cdot{\frac{{{1}}}{{{4}^{{{2}}}}}}-{\frac{{{1}}}{{{1}^{{{2}}}}}}\)
\(\displaystyle{\frac{{{1}}}{{\lambda}}}={97.2}{n}{m}\)
Therefore
The Correct slots are
\(\displaystyle{n}{f}={2},{n}{i}={1}\Rightarrow{121.5}{n}{m}\)
\(\displaystyle{n}{i}={1},{n}{f}={3}\Rightarrow{102.6}{n}{m}\)
\(\displaystyle{n}{i}={1},{n}{f}={4}\Rightarrow{97.2}{n}{m}\)
0
star233
Answered 2022-01-15 Author has 0 answers

Step 1
Given:
\(\\\lambda_{1} = 121.5 nm \\\lambda_{2} = 102.6 nm \\\lambda_{3} = 97.23 nm\)
Concluding that these wavelengths are shorter than that of visible light. Therefore, assuming the final energy level, nf to be 1.
The energy change and wavelength are related as:
\(\triangle E=-\frac{hc}{\lambda}\)
Step 2
Therefore, calculating the energies associated with the given wavelengths using the above equation:
\(\\\triangle E_{1}=-\frac{6.626 \times 10^{-34}J.s \times 3.00 \times 10^{8}ms^{-1}}{121.5 \times 10^{-9}m} \\=-1.64 \times 10^{-18}J \\\triangle E_{2}=-\frac{6.626 \times 10^{-34}J.s \times 3.00 \times 10^{8}ms^{-1}}{102.6 \times 10^{-9}m} \\=-1.94 \times 10^{-18}J \\\triangle E_{3}=-\frac{6.626 \times 10^{-34}J.s \times 3.00 \times 10^{8}ms^{-1}}{97.23 \times 10^{-9}m} \\=-2.045 \times 10^{-18}J\)
Rydberg's formula is given as:
Z - Atomic no. which for H-atom is 1
R - Rydberg's constant \(= 2.18 \times 10-18 J\)
Step 3
Therefore, finding the initial energy levels associated with each of these wavelengths as:
Case 1:
\(\\\triangle E_{1}=-2.18 \times 10^{-18}J [\frac{1}{(1)^{2}}-\frac{1}{n_{i}^{2}}] \\ [\frac{1}{(1)^{2}}-\frac{1}{n_{i}^{2}}]=\frac{\triangle E_{1}}{-2.18 \times 10^{-18}J} \\\frac{1}{n_{i}^{2}}=\frac{1}{(1)^{2}}+\frac{\triangle E_{1}}{2.18 \times 10^{-18}J} \\n_{i}=\sqrt{\frac{1}{\frac{1}{(1)^{2}}+\frac{\triangle E_{1}}{2.18 \times 10^{-18}J}}} \\=\sqrt{\frac{1}{\frac{1}{(1)^{2}}+\frac{-1.64 \times 10^{-18}J}{2.18 \times 10^{-18}J}}} \\=\sqrt{\frac{1}{0.247}} \\=\sqrt{4.04} \\=2.01 \\\approx 2\)
Case 2:
\(\triangle E_{1}=-2.18 \times 10^{-18}J[\frac{1}{(1)^{2}}-\frac{1}{n_{i}^{2}}] \\ [\frac{1}{(1)^{2}}-\frac{1}{n_{i}^{2}}]=\frac{\triangle E_{1}}{-2.18 \times 10^{-18}J} \\\frac{1}{n_{i}^{2}}=\frac{1}{(1)^{2}}+\frac{\triangle E_{1}}{2.18 \times 10^{-18}J} \\n_{i}=\sqrt{\frac{1}{\frac{1}{(1)^{2}}+\frac{\triangle E_{1}}{2.18 \times 10^{-18}J}}} \\=\sqrt{\frac{1}{\frac{1}{(1)^{2}}+\frac{-1.94 \times 10^{-18}J}{2.18 \times 10^{-18}J}}} \\=\sqrt{\frac{1}{0.11}} \\=\sqrt{9.09} \\=3.01 \\\approx 3\)
Case 3:
\(\triangle E_{1}=-2.18 \times 10^{-18}J [\frac{1}{(1)^{2}}-\frac{1}{n_{i}^{2}}] \\ [\frac{1}{(1)^{2}}-\frac{1}{n_{i}^{2}}]=\frac{\triangle E_{1}}{-2.18 \times 10^{-18}J} \\\frac{1}{n_{i}^{2}}=\frac{1}{(1)^{2}}+\frac{\triangle E_{1}}{2.18 \times 10^{-18}J} \\n_{i}=\sqrt{\frac{1}{\frac{1}{(1)^{2}}+\frac{\triangle E_{1}}{2.18 \times 10^{-18}J}}} \\=\sqrt{\frac{1}{\frac{1}{(1)^{2}}+\frac{-2.045 \times 10^{-18}J}{2.18 \times 10^{-18}J}}} \\=\sqrt{\frac{1}{0.062}} \\=\sqrt{16.13} \\=4.01 \\\approx 4\)
Therefore, the transitions take place as:
\(\lambda_{1}:2 \Rightarrow 1 \\\lambda_{2}:3 \Rightarrow 1 \\\lambda_{3}:4 \Rightarrow 1\)

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