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A photon has a frequency of \(6.0 \times 10^{4} \mathrm{~Hz}\). (a) Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? (b) Calculate the energy (in joules) of this photon. (c) Calculate the energy (in joules) of 1 mole of photons all with this frequency.

Short Answer

Expert verified
The wavelength of the photon is \(5 \times 10^{9} nm\), not within the visible light spectrum. The energy of a single photon is \(3.9756 × 10^{-29} Joules\), and one mole of these photons have energy \(2.3922 Joules\)

Step by step solution

01

Convert Frequency to Wavelength

To solve this part, rearrange \( c = \lambda v \) for the wavelength. That results in \( \lambda = \frac{c}{v} \). Now plug in the given values: speed of light \( c = 3.0 \times 10^{8} m/s \) and frequency \( v = 6.0 \times 10^{4} Hz \). So \( \lambda = \frac{3.0 \times 10^{8} m/s}{6.0 \times 10^{4} Hz} = 5000 m = 5 \times 10^{9} nm \). The visible light spectrum ranges from 400-700 nm, so this frequency do not falls in the visible region.
02

Calculate the Energy of a Photon

Now using the equation \( E = h v \) where \( h = 6.626 × 10^{-34} Js \) is the Planck’s constant and frequency \( v = 6.0 \times 10^{4} Hz \). Calculate the energy of a single photon by substituting the given values into the equation: \( E = 6.626 × 10^{-34} Js \times 6.0 \times 10^{4} Hz = 3.9756 × 10^{-29} Joules \).
03

Calculate the Energy of One Mole of Photons

Finally, to find the energy of 1 mole of these photons, multiply the energy of one photon by Avogadro's number (which is approximately \(6.022 \times 10^{23}\) particles/mole). So, \( E_{mole} = E_{photon} \times N_{Avogadro} = 3.9756 × 10^{-29} Joules \times 6.022 \times 10^{23} mole^{-1} = 2.3922 Joules \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wavelength and Frequency Conversion
Understanding the relationship between a photon's frequency and its wavelength is crucial in various fields of science, especially in physics and chemistry. This relationship is dictated by the wave equation, which connects the speed of light (\( c \)), frequency (\( v \)), and wavelength (\( \lambda \) ) in the formula: \[\lambda = \frac{c}{v}\]. The speed of light (\( c \) is a constant, approximately \(3 \times 10^{8} \text{m/s}\), and applying this equation, you can easily convert between the frequency and wavelength of electromagnetic waves, including visible light. For example, the frequency of \(6.0 \times 10^{4} \text{Hz}\) converted to wavelength in nanometers (\(nm\)) doesn't fall within the visible spectrum range of 400-700 nm, indicating that it's not light we can see with the naked eye. The reason it's essential to learn and understand this conversion process is because it allows scientists and students to describe and categorize different types of electromagnetic radiation — from radio waves to gamma rays — based on their wavelengths and frequencies.

Additionally, this conversion is fundamental in technologies like lasers, telecommunications, and medical imaging, where precise control of wavelength and frequency is necessary for proper functionality.
Planck's Constant Application
Planck's constant (\( h \) is a fundamental constant in quantum mechanics and plays a pivotal role in the energy quantization of photons. Representing the smallest quantity of energy that can be emitted or absorbed as electromagnetic radiation, Planck’s constant is approximately \(6.626 \times 10^{-34} \text{Js}\). When it comes to calculating the energy (\( E \) of a photon, the equation \[\ E = h v \] comes into play, where \( v \) is the frequency of the photon.

This equation suggests that the energy of a photon is directly proportional to its frequency, and understanding this concept is essential in fields like spectroscopy, photochemistry, and even the development of solar panels. For example, a photon with the frequency of \(6.0 \times 10^{4} \text{Hz}\) will have an energy of \(3.9756 \times 10^{-29} \text{Joules}\), calculated using Planck's constant. Students should grasp this concept to understand the behaviors of photons under different conditions, which is pivotal in studies concerning the nature of light and its interaction with matter.
Avogadro's Number in Chemistry
In the field of chemistry, Avogadro's number — approximately \(6.022 \times 10^{23} \text{particles/mole}\) — represents the number of constituent particles, usually atoms or molecules, in one mole of a substance.

This incredibly large number is paramount in calculations that involve converting between the number of particles and the amount of substance in moles. For instance, knowing the energy of a single photon, as calculated with Planck's constant, allows us to calculate the energy contained in one mole of identical photons by multiplying the energy of one photon by Avogadro's number. Such calculations are key in understanding mole concepts and are utilized to determine the amount of energy transferable in chemical reactions or materials, which has implications for fields such as pharmacology, material science, and energy production.

As seen in the exercise, 1 mole of the given photons with a frequency of \(6.0 \times 10^{4} \text{Hz}\) would have a total energy of \(2.3922 \text{Joules}\), illustrating the collective impact of this number of photons and explaining why Avogadro's number is not only a fundamental constant in chemistry but also in physics and other sciences that deal with particulate matter.

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Most popular questions from this chapter

The \(\mathrm{He}^{+}\) ion contains only one electron and is therefore a hydrogenlike ion. Calculate the wavelengths, in increasing order, of the first four transitions in the Balmer series of the \(\mathrm{He}^{+}\) ion. Compare these wavelengths with the same transitions in a \(\mathrm{H}\) atom. Comment on the differences. (The Rydberg constant for \(\mathrm{He}^{+}\) is \(\left.8.72 \times 10^{-18} \mathrm{~J} .\right)\)

In a photoelectric experiment a student uses a light source whose frequency is greater than that needed to eject electrons from a certain metal. However, after continuously shining the light on the same area of the metal for a long period of time the student notices that the maximum kinetic energy of ejected electrons begins to decrease, even though the frequency of the light is held constant. How would you account for this behavior?

In the beginning of the twentieth century, some scientists thought that a nucleus may contain both electrons and protons. Use the Heisenberg uncertainty principle to show that an electron cannot be confined within a nucleus. Repeat the calculation for a proton. Comment on your results. Assume the radius of a nucleus to be \(1.0 \times 10^{-15} \mathrm{~m} .\) The masses of an electron and a proton are \(9.109 \times 10^{-31} \mathrm{~kg}\) and \(1.673 \times 10^{-27} \mathrm{~kg},\) respectively. (Hint: Treat the diameter of the nucleus as the uncertainty in position.)

Give the values of the four quantum numbers of an electron in the following orbitals: (a) \(3 s,\) (b) \(4 p\), (c) \(3 d\).

Scientists have found interstellar hydrogen atoms with quantum number \(n\) in the hundreds. Calculate the wavelength of light emitted when a hydrogen atom undergoes a transition from \(n=236\) to \(n=235 .\) In what region of the electromagnetic spectrum does this wavelength fall?

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