2.1: Question 2.E.06 PASS - Energy, wavelength, frequency, and colour of emitted lithium photons

Exercise \(\PageIndex{2.E.06}\)

Heated lithium atoms emit photons of light with an energy of 2.961 ×10⁻¹⁹ J. Calculate the frequency and wavelength of one of these photons. What is the total energy in 1 mole of these photons? What is the color of the emitted light?

Answer

\begin{aligned} & \text { Frequency }(\mathrm{v})=4.469 \times 10^{14} \mathrm{~s}^{-1} \\ & \text { Wavelength }(\lambda)=6.713 \times 10^{-7} \mathrm{~m} \\ & \text { Energy in } 1 \text { mole }=1.783 \times 10^5 \mathrm{~J} \\ & \text { Colour of light }=\text { Red }\end{aligned}

See LibreText Section 2.1 Electromagnetic Energy

See LibreText Example 2.1.2 Calculating the Energy of Radiation

Strategy Map

Step	Hint
1. Identify the information provided in the question.	Recall: LibreText Example 2.1.2 Calculating the Energy of Radiation
2. Choose a formula that connects the information that is provided and the frequency and wavelength.	You will need to use two formulas
3. Using provided data to calculate the amount of energy in one mole.	How many “items” are in one mole? How can you convert a photon to a mole?
4. Compare your wavelength to the electromagnetic spectrum, what colour would your calculated wavelength fall under?	Try converting your wavelength into nanometers.

Solution

Frequency of emitted photon

\begin{gathered}E=h v \\ 2.691 \times 10^{-19} J=\left(6.626 \times 10^{-34} J s\right) v \\ v=\frac{6.691 \times 10^{-19} J}{6.626 \times 10^{-34} J s} \\ v=\mathbf{4 . 4 6 9} \times \mathbf{1 0}^{\mathbf{1 4}} \mathbf{s}^{-1}\end{gathered}

Wavelength of emitted photon

\begin{gathered}c=\lambda v \\ 3.00 \times 10^8 \mathrm{~m} / \mathrm{s}=\left(4.469 \times 10^{14} \mathrm{~s}^{-1}\right) \lambda \\ \lambda=\frac{3.00 \times 10^8 \mathrm{~m} / \mathrm{s}}{4.469 \times 10^{14} \mathrm{~s}^{-1}} \\ \lambda=\mathbf{6 . 7 1 3 \times 1 0 ^ { - 7 } \mathrm { m }}\end{gathered}

Energy in 1 mole of emitted photons

\begin{gathered}\frac{2.961 \times 10^{-19} \mathrm{~J}}{1 \text { photon }} \times \frac{6.022 \times 10^{23} \text { photons }}{1 \text { mole }} \\ =\mathbf{1 . 7 8 3 \times 1 0 ^ { 5 } \mathrm { J } / \mathbf { m o l }}\end{gathered}

Colour of emitted light

\begin{gathered}6.713 \times 10^{-7} \mathrm{~m} \times \frac{1 \times 10^9 \mathrm{~nm}}{1 \mathrm{~m}}=671 \mathrm{~nm} \\ =\text { Red Light }\end{gathered}

Guided Solution

Download Guided Solution as a pdf

Guided Solution	Hint
This is a multiple part calculation question where you use related equations to solve for frequency and wavelength. Next to determine secondary information use constant values and to calculate the energy in one mole of the given photons and conversion factors to predict the colour of light emitted.	Refer to: LibreText Example 2.1.2 Calculating the Energy of Radiation
Highlight important information provided in the question: Heated lithium atoms emit photons of light with an energy of 2.961 ×10^−19 J. Calculate the frequency and wavelength of one of these photons. What is the total energy in 1 mole of these photons? What is the color of the emitted light?”	When reading through a question to know what to find look for key words such as “Calculate” and phrases such as “What is”. - “Calculate the frequency and wavelength of one of these photons.” - “What is the total energy in 1 mole of these photons?” - “What is the color of the emitted light?”
An example of the same question with key information emphasized: Heated lithium atoms emit photons of light with an energy of 2.961 × 10^−19 J. Calculate the frequency and wavelength of one of these photons. What is the total energy in 1 mole of these photons? What is the color of the emitted light?	A summary of the key information provided: What we know: We know that the energy (E) of an emitted photon is equal to 2.961 × 10^−19 J. What we are asked: We are asked to find the frequency (v), wavelength (λ), the energy in one mole, and the colour of the emitted light.
What else do you know about this topic? Avogadro's Number defines the number of 'items' in a mole Visual Light spectrum: Violet: 400-420 nm Indigo: 420-440 nm Blue: 440-490 nm Green: 490-570 nm Yellow: 570-585 nm Orange: 585-620 nm Red: 620-780 nm	Recall that you know how many particles are in one mole. (Avogadro's Number = 2.066x10^23) Photons are also known as particles. We can use this value in a conversion calculation to convert our photons into one mole of photons. Recall that each colour on the light spectrum is seen as a different wavelength, typically seen in nanometers. To find the colour of light convert your wavelength into nanometers.
Complete Solution: Step 1 Using Planck’s equation, we can rearrange and solve for the wave’s frequency (v). We are given the energy and we know Planck’s constant (h = 6.626x10^-34 Js) \begin{gathered}E=h v \\ 2.691 \times 10^{-19} J=\left(6.626 \times 10^{-34} J s\right) v \\ v=\frac{6.691 \times 10^{-19} J}{6.626 \times 10^{-34} J s} \\ v=\mathbf{4 . 4 6 9} \times \mathbf{1 0}^{\mathbf{1 4}} \mathbf{s}^{-1}\end{gathered} Step 2 We can now rearrange and solve for our wavelength (λ) using the frequency we just found and the speed of light (c = 3.00x10^8 m/s) in the equation c = λv. \begin{gathered}c=\lambda v \\ 3.00 \times 10^8 \mathrm{~m} / \mathrm{s}=\left(4.469 \times 10^{14} \mathrm{~s}^{-1}\right) \lambda \\ \lambda=\frac{3.00 \times 10^8 \mathrm{~m} / \mathrm{s}}{4.469 \times 10^{14} \mathrm{~s}^{-1}} \\ \lambda=\mathbf{6 . 7 1 3 \times 1 0 ^ { - 7 } \mathrm { m }}\end{gathered} Step 3 To find the energy in one mole of these photons we need to use Avogadro's number as a conversion factor. There are 2.066x10^23 photons in one mole. \begin{gathered}\frac{2.961 \times 10^{-19} \mathrm{~J}}{1 \text { photon }} \times \frac{6.022 \times 10^{23} \text { photons }}{1 \text { mole }} \\ =\mathbf{1 . 7 8 3 \times 1 0 ^ { 5 } \mathrm { J } / \mathbf { m o l }}\end{gathered} Step 4 To find the colour of light emitted, first convert from meters to nanometers (this makes it easier to compare the calculated value to known colour ranges in the visible light spectrum). \begin{gathered}6.713 \times 10^{-7} \mathrm{~m} \times \frac{1 \times 10^9 \mathrm{~nm}}{1 \mathrm{~m}}=671 \mathrm{~nm} \\ =\text { Red Light }\end{gathered} The calculated value falls in the range of red light which is between 620 and 780 nm.	Re: Step 3 - To set up a conversion properly, be sure that the numerator has the unit you want, and the denominator has the unit that you must cancel out. Re: Step 4 - Recall that each colour on the light spectrum has a different wavelength, typically described in nanometers . To find the colour of light convert your wavelength into nanometers. There are 10^9 nanometers in 1 meter; alternately 1 nanometer is 10^-9 meters. Either conversion factor will work, just be sure that the units cancel out appropriately. Violet: 400-420 nm; Indigo: 420-440 nm; Blue: 440-490 nm; Green: 490-570 nm; Yellow: 570-585 nm; Orange: 585-620 nm; Red: 620-780 nm
Check your work! Why does this answer make chemical sense? Wavelength and frequency are inversely proportional to one another, as frequency increases, wavelength decreases. This is demonstrated in our answer as the wavelength is very small and the frequency is very large. The colour of light is dependent on the wavelength of the traveling light. Different wavelengths will be absorbed and reflected differently producing different colours in the visible light spectrum. In this case the wavelength was on the higher end of that spectrum and thus produced red light.

(question source from LibreText page titled 6.E:Electronic Structure and Periodic Properties (Exercises): https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/06%3A_Electronic_Structure_and_Periodic_Properties_of_Elements/6.E%3A_Electronic_Structure_and_Periodic_Properties_(Exercises), shared under a CC BY 4.0 license, authored, remixed, and/or curated by OpenStax, original source https://openstax.org/books/chemistry-2e/pages/6-exercises , Access for free at https://openstax.org/books/chemistry/pages/1-introduction)