Question

Consider the following approximate visible light spectrum: Barium emits light in the visible region of the spectrum. If each photon of light emitted from barium has an energy of $3.59\times {10}^{-19}\mathrm{J},$ what color of visible light is emitted?

Short answer

The emitted light color can be determined by calculating the wavelength of the light and comparing it with the known visible light spectrum. Using Planck's energy formula and the speed of light equation, we find that the wavelength of the emitted light is $\lambda =554\mathrm{nm}$. Since this falls within the green color range (495-570 nm) in the visible light spectrum, the emitted light is green.

Step-by-step solution

Recall the Planck's energy formula

Planck's energy formula relates the energy of a photon to its frequency and wavelength. The formula is given by: $$E=h\times \nu$$ where $E$ represents the energy of the photon, $h$ is the Planck's constant $(6.63\times {10}^{-34}\mathrm{Js})$, and $\nu$ is the frequency of the light. We also know the speed of light in vacuum $c=3.00\times {10}^8\mathrm{m}/\mathrm{s}$ and the relationship between the speed of light, frequency, and wavelength: $$c=\nu \times \lambda$$ where $\lambda$ is the wavelength of the light.

Calculate the frequency of the light

Using the given information, first, we need to find the frequency of the light. Using the energy formula, we can rewrite the equation for frequency as: $$\nu =\frac{E}{h}$$ Now, substitute the given energy value $E=3.59\times {10}^{-19}\mathrm{J}$ and Planck's constant $h=6.63\times {10}^{-34}\mathrm{Js}$ into the formula: $$\nu =\frac{3.59\times {10}^{-19}\mathrm{J}}{6.63\times {10}^{-34}\mathrm{Js}}$$

Solve for frequency

By calculating the value, we get the frequency: $$\nu =5.41\times {10}^{14}\mathrm{Hz}$$

Calculate the wavelength of the light

Next, we will calculate the wavelength of the light using the speed of light equation: $$\lambda =\frac{c}{\nu }$$ Substitute the calculated frequency value $\nu =5.41\times {10}^{14}\mathrm{Hz}$ and the speed of light constant $c=3.00\times {10}^8\mathrm{m}/\mathrm{s}$ into the formula: $$\lambda =\frac{3.00\times {10}^8\mathrm{m}/\mathrm{s}}{5.41\times {10}^{14}\mathrm{Hz}}$$

Solve for the wavelength of the light

By calculating the values, we get the wavelength: $$\lambda =5.54\times {10}^{-7}\mathrm{m}$$ or $$\lambda =554\mathrm{nm}$$

Determine the color of the light

Now, we can compare the calculated wavelength with the known visible light spectrum to find the corresponding color. Visible light wavelengths are typically categorized as follows: - Violet: 380-450 nm - Blue: 450-495 nm - Green: 495-570 nm - Yellow: 570-590 nm - Orange: 590-620 nm - Red: 620-750 nm Since the calculated wavelength is 554 nm, which falls within the green color range, the emitted light is green.

Key concepts

Planck's Energy Formula

Understanding Planck's energy formula is essential when exploring the fundamental properties of light. At the heart of quantum theory, this formula connects the energy of a photon to its frequency. It is succinctly represented by the equation:
$$E=h\times u$$
Here, E stands for the energy of the photon, h is a physical constant known as Planck's constant (exactly $6.62607015\times {10}^{-34}$ joule seconds), and u (nu) symbolizes the frequency of the photon.
By acknowledging that the frequency u is the number of waves that pass a given point every second, we deduce that higher frequency light has more energy per photon. This understanding is pivotal in fields ranging from astronomy to communications technology, where the energy of light dictates how we can use and interpret it.

Application in Calculating Photon Energy

Planck's formula becomes particularly useful when you aim to find the energy of light, like a photon emitted by barium in our example exercise. Knowing the energy allows us to further deduce properties such as the light's wavelength and subsequently its color within the visible spectrum—details key for applications in spectroscopy and optical engineering.

Photon Energy

The concept of photon energy relates to the quantized nature of light. Photons are elemental particles of light, each harboring a distinct packet of energy. This energy, as we mentioned, can be expressed using Planck's energy formula.
Photon energy holds the key to not only understanding light's color but also its potential to facilitate or initiate various processes. For instance, a photon's energy must meet or exceed certain thresholds to affect chemical reactions, such as photosynthesis in plants or the production of vitamin D in human skin.

Seeing the Energy

When discussing visible light specifically, the energy of a photon can determine what color we perceive. Higher energy photons correspond to colors towards the violet end of the spectrum, while lower energy photons are seen as red. This is because our eyes have different receptors for colors that get activated when struck by photons of specific energies. By employing the provided energy information of a barium-emitted photon and Planck's constant, we could infer the frequency and thus the color of light in the exercise.

Light Wavelength and Color

The concept of light wavelength is intimately tied to the perception of color in the visible light spectrum. Wavelength is the distance over which the wave's shape repeats, and it is usually measured in nanometers (nm) when it comes to light.
The visible spectrum is a small window of wavelengths that human eyes can detect, usually ranging from about 380 nm to 750 nm. Each range of wavelengths corresponds to a specific color that we can see. Violet light has the shortest wavelength, while red has the longest.

From Wavelength to Color Perception

Taking our barium photon with an energy of $3.59\times {10}^{-19}$ joules, we calculated the wavelength to be around 554 nm. This calculation places it within the green region of the visible spectrum, which is why we would see the emitted light as green. This is how, by knowing the energy of a photon or its wavelength, you can deduce the color of the light—an essential aspect in applications ranging from art conservation to the design of LED lighting.