Question

When a compound containing cesium ion is heated in a Bunsen burner flame, photons with an energy of \(4.30 \times 10^{-19} \mathrm{~J}\) are emitted. What color is the cesium flame?????

Short answer

The cesium flame color is blue.

Step-by-step solution

Determine Frequency Using Energy

Given the energy of the photons emitted as \( E = 4.30 \times 10^{-19} \text{ J} \), we use the equation for energy of a photon, which is \( E = h \cdot f \), to find the frequency \( f \). Here, \( h \) is Planck's constant, \( 6.63 \times 10^{-34} \text{ Js} \). Rearrange the equation to solve for \( f \): \( f = \frac{E}{h} \). Substitute the values:\[ f = \frac{4.30 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 6.48 \times 10^{14} \text{ Hz} \].

Determine Wavelength from Frequency

Now that we have the frequency, we use the equation \( c = \lambda \cdot f \) to find the wavelength \( \lambda \). Here, \( c \) is the speed of light, approximately \( 3.00 \times 10^8 \text{ m/s} \). Rearrange the equation to solve for \( \lambda \): \( \lambda = \frac{c}{f} \). Substitute the frequency:\[ \lambda = \frac{3.00 \times 10^8}{6.48 \times 10^{14}} \approx 4.63 \times 10^{-7} \text{ m} \].

Convert Wavelength to Nanometers

Since it is common to express wavelength in nanometers (nm) rather than meters (m), we convert the result from meters to nanometers. \( 1 \text{ m} = 10^9 \text{ nm} \), so we multiply by \( 10^9 \): \( 4.63 \times 10^{-7} \text{ m} = 463 \text{ nm} \).

Identify the Flame Color from Wavelength

The visible light spectrum can be used to identify colors associated with specific wavelengths. A wavelength of 463 nm falls in the blue range of the visible light spectrum, so the color of the flame is blue.

Key concepts

Photon Energy

Photon energy is a crucial concept in understanding many aspects of light and electromagnetic radiation. In simple terms, the energy of a photon is the amount of energy carried by a single particle of light. This energy can be calculated using the formula:
\[ E = h \cdot f \]
where:

• \(E\) is the photon energy in joules (J)
• \(h\) is Planck's constant, \(6.63 \times 10^{-34} \text{ Js}\)
• \(f\) represents the frequency of the light in hertz (Hz)

The energy of a photon is directly proportional to its frequency. Higher frequency light, like ultraviolet, carries more energy than lower frequency light like infrared. This concept is essential in explaining how different colors of light possess different energies. When heating cesium ions, the emission of photons at a certain energy level can be related to the specific color of light observed in the flame test.

Wavelength Calculation

Wavelength calculation is another important part of understanding light emission. In the context of cesium flame tests, once you know the frequency of a photon, you can calculate its wavelength. The formula used is:
\[ c = \lambda \cdot f \]
where:

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

By rearranging the formula to solve for the wavelength, \(\lambda\), you get:
\[ \lambda = \frac{c}{f} \]
This provides the length, in meters, of one complete wave cycle of the light being emitted. Understanding this relationship helps explain why different elements can produce different flame colors, as each emits photons of distinctive frequencies and, hence, wavelengths.

Visible Light Spectrum

The visible light spectrum is a small part of the electromagnetic spectrum that the human eye can perceive. It ranges approximately from 380 nm to 750 nm in wavelength. Depending on the wavelength of light, different colors are perceived.

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

In the cesium flame test, photons are emitted with a wavelength of 463 nm, placing them in the blue portion of the spectrum. As such, you observe a blue flame. Understanding this spectrum is key in identifying unknown elements based on flame color.

Planck's Constant

Planck's constant is a fundamental constant in physics, critical to the study of quantum mechanics and electromagnetic phenomena. It relates the energy carried by a photon to its frequency. Planck’s constant (\(h\)) is approximately\(6.63 \times 10^{-34} \text{ Js}\).
This constant allows us to bridge the gap between macroscopic observations of light and their quantum mechanical explanations. It underscores how light behaves as particles, allowing us to calculate specific energies of the photons emitted during various processes, such as the heating of cesium ions.
With Planck's constant, we can grasp the dual wave-particle nature of light, which is central to quantum theory. This not only illuminates the behavior of light but also provides insight into how atoms and particles interact with electromagnetic radiation. By understanding Planck’s constant and its application, we can explain diverse phenomena including color emissions, as observed in flame tests.