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 color of the visible light emitted from barium with an energy of \(3.59 \times 10^{-19} \mathrm{~J}\) per photon is green. We found this by calculating the frequency of the photon (\(5.42 \times 10^{14} \mathrm{~Hz}\)) and the wavelength (\(5.53 \times 10^{-7} \mathrm{~m}\) or \(553 \mathrm{~nm}\)), which falls within the green color range in the visible light spectrum.

Step-by-step solution

Recall the required formulas

To solve this problem, we need to recall the formula for the energy of a photon, given by: \(E = h \cdot f\), where: - \(E\) is the energy of the photon, - \(h\) is the Planck constant (\(6.63 \times 10^{-34} \mathrm{~Js}\)), - \(f\) is the frequency of the light wave. Additionally, we need to recall the relationship between frequency \(f\) and wavelength \(\lambda\) of a light wave, which is given by the speed of light formula: \(\lambda = \frac{c}{f}\), where: - \(\lambda\) is the wavelength, - \(c\) is the speed of light (\(3.00 \times 10^8 \mathrm{~m/s}\)).

Calculate the frequency of the photon

Firstly, we should calculate the frequency \(f\) using the given energy \(E\) of the photon. We can use the energy formula, \(E = h \cdot f\). Rearranging for frequency, we get: \(f = \frac{E}{h}\), Now, substitute the given values to find the frequency: \(f = \frac{3.59 \times 10^{-19} \mathrm{~J}}{6.63 \times 10^{-34} \mathrm{~Js}} \approx 5.42 \times 10^{14} \mathrm{~Hz}\).

Calculate the wavelength of the photon

With the frequency calculated, we can now determine the wavelength of the light using the speed of light formula \(\lambda = \frac{c}{f}\). Substitute the values obtained in Steps 1 and 2: \(\lambda = \frac{3.00 \times 10^8 \mathrm{~m/s}}{5.42 \times 10^{14} \mathrm{~Hz}} \approx 5.53 \times 10^{-7} \mathrm{~m}\).

Determine the color of the light wave

Using the wavelength obtained in Step 3, we can now identify the corresponding color in the visible light spectrum. The wavelengths of the visible spectrum approximately fall within the following range: - Violet: \(380 - 450 \mathrm{~nm}\), - Blue: \(450 - 495 \mathrm{~nm}\), - Green: \(495 - 570 \mathrm{~nm}\), - Yellow: \(570 - 590 \mathrm{~nm}\), - Orange: \(590 - 620 \mathrm{~nm}\), - Red: \(620 - 750 \mathrm{~nm}\). Since we obtained a wavelength of approximately \(5.53 \times 10^{-7} \mathrm{~m}\), or \(553 \mathrm{~nm}\), the emitted light falls within the green color range. Hence, the color of the visible light emitted is green.

Key concepts

Visible Light Spectrum

The visible light spectrum is a part of the electromagnetic spectrum that is visible to the human eye. It encompasses a range of colors, each defined by distinct wavelengths. In simple terms, when lights of different wavelengths enter our eye, they are perceived as different colors. The visible spectrum includes the familiar rainbow of colors:

• Violet: 380 - 450 nm
• Blue: 450 - 495 nm
• Green: 495 - 570 nm
• Yellow: 570 - 590 nm
• Orange: 590 - 620 nm
• Red: 620 - 750 nm

The units used for wavelengths in these ranges are nanometers (nm), where 1 nm equals 1 billionth of a meter. When we refer to the color green, it has a wavelength of approximately 495 to 570 nm. This range represents a specific part of the visible light spectrum.

Planck Constant

The Planck constant is essential in the field of quantum mechanics. It describes the size of the energy quanta emitted or absorbed by matter. Denoted by the symbol \(h\), its approximate value is \(6.63 \times 10^{-34} \text{ Js}\). This small constant plays a crucial role in calculating photon energy, linking the energy of a photon to its frequency with the formula:\[E = h \cdot f\]Here:

• \(E\) is the energy of a photon (in joules).
• \(f\) is the frequency of the light wave (in hertz).

The Planck constant allows us to bridge the gap between the macroscopic observations in physics and the microscopic domain where quantum phenomena occur. It's invaluable for calculating energy at the quantum level.

Wavelength Calculation

Calculating the wavelength of a light wave is a critical step in determining its color within the visible spectrum. Once the frequency \(f\) of the light is known, the wavelength \(\lambda\) can be calculated using the speed of light \(c\) and this formula:\[\lambda = \frac{c}{f}\]In this formula:

• \(\lambda\) represents the wavelength (in meters).
• \(c\) is the speed of light, approximately \(3.00 \times 10^8 \text{ m/s}\).

Suppose you have a wavelength of \(553\) nm (or \(5.53 \times 10^{-7} \) m) for a light wave. This places the light within the green part of the visible spectrum. Understanding wavelength allows us to interpret different lights' properties and connect them to specific aspects of the electromagnetic spectrum.

Frequency of Light

The frequency of light, denoted as \(f\), describes how many waves pass a specific point in one second. It is measured in hertz (Hz). The relationship between frequency and wavelength is inversely proportional, governed by the speed of light. To find the frequency of a photon, given its energy \(E\), use the formula:\[f = \frac{E}{h}\]This formulation requires:

• The photon energy \(E\) in joules.
• The Planck constant \(h\), which is \(6.63 \times 10^{-34} \text{ Js}\).

For example, by knowing a photon energy of \(3.59 \times 10^{-19} \text{ J}\), you calculate its frequency as approximately \(5.42 \times 10^{14} \text{ Hz}\). This frequency value then assists in determining the wavelength and consequently the color of the light. Recognizing how frequency translates to observable characteristics illuminates remarkable aspects of light behavior.