Problem 25 - Chemistry The Molecular Science

Question

Flame tests depend on emissions in the visible region of the spectrum to identify elements in a sample. Bariumcontaining compounds emit at \(493 \mathrm{nm}\); strontiumcontaining compounds emit at \(642 \mathrm{nm}\). (a) Determine the identifying color of the flame test in each case. (b) Calculate the energy (J/photon) associated with each of these emissions. (c) Explain why the barium emission occurs at lower wavelength than the strontium does.

Step-by-Step Solution

Answer

Barium: blue-green, approx. 4.03 x 10^-19 J/photon; Strontium: red, approx. 3.10 x 10^-19 J/photon. Barium emits at a shorter wavelength, meaning higher energy.

Step 1: Identify Flame Colors

To identify the flame colors, we use the wavelength of the emission. Barium compounds emit at \(493 \ \text{nm}\), which corresponds to a blue-green color. Strontium compounds emit at \(642 \ \text{nm}\), which corresponds to a red color.

Step 2: Calculate Energy of Photons (Barium)

The energy of a photon is calculated using the formula \(E = \frac{hc}{\lambda}\), where \(h = 6.626 \times 10^{-34} \ \text{J}\cdot\text{s}\) is Planck's constant and \(c = 3 \times 10^8 \ \text{m/s}\) is the speed of light. For barium's emission at \(493 \ \text{nm} = 493 \times 10^{-9} \ \text{m}\), the energy is \[E = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{493 \times 10^{-9}}\] which calculates to approximately \(4.03 \times 10^{-19} \ \text{J/photon}\).

Step 3: Calculate Energy of Photons (Strontium)

Using the same formula for energy, substitute strontium's emission wavelength, \(642 \ \text{nm} = 642 \times 10^{-9} \ \text{m}\). The energy is \[E = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{642 \times 10^{-9}}\] which calculates to approximately \(3.10 \times 10^{-19} \ \text{J/photon}\).

Step 4: Compare Wavelength and Energy

The barium emission occurs at a shorter wavelength (493 nm) than strontium (642 nm). According to the relation between wavelength and energy (\(E = \frac{hc}{\lambda}\)), shorter wavelengths correspond to higher energy. Hence, barium's emission has higher energy than strontium's.

Key Concepts

Emission Wavelength

When elements are heated, they emit light. This light has specific wavelengths that are unique to each element, similar to a fingerprint. The flame test utilizes these emissions to identify the presence of an element in a sample. Barium compounds emit at a wavelength of 493 nm, which falls within the blue-green region of the visible spectrum. Conversely, strontium compounds emit at 642 nm, which is in the red portion.

This difference in emission wavelengths is due to variations in each element’s electron arrangement and energy transitions.

• Shorter wavelengths correspond to blue and green colors.
• Longer wavelengths are linked to red and orange colors.

Therefore, identifying elements based on their flame color during a flame test provides a visual representation of these specific wavelengths.

Photon Energy Calculation

Photon energy is an essential concept when understanding emissions in flame tests. The energy of a photon is determined using the formula: \[E = \frac{hc}{\lambda}\] where:

• h is Planck’s constant:

  6.626 x 10^-34 J·s

• c is the speed of light:

  3 x 10⁸ m/s

• λ (lambda) is the wavelength:

  measured in meters.

For barium's emission at 493 nm (which converts to 493 x 10^-9 m), the calculation is approximately 4.03 x 10^-19 J/photon. Strontium's longer wavelength of 642 nm yields a lower photon energy of around 3.10 x 10^-19 J/photon.

The key takeaway is that shorter wavelengths provide higher energy, while longer wavelengths result in lower energy emissions.

Visible Spectrum Emissions

The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye. It ranges approximately from 400 nm (violet) to 700 nm (red). In the context of flame tests, when compounds are heated, they emit light within this range due to the movement of electrons between energy levels.

This emitted light allows the identification of different elements:

• Light from 400-500 nm appears blue or green.
• Light from 600-700 nm looks red.

Since barium and strontium emit light at 493 nm and 642 nm respectively, they are observable as blue-green and red colors. These emissions are crucial for identifying and differentiating elements based on their unique spectral emissions.

Understanding visible spectrum emissions solidifies one's grasp of why different elements display distinct colors during a flame test, serving as their spectral signature.