Question

The flame test for barium involves placing a barium-containing compound in a Bunsen burner flame. The barium compound then decomposes to give barium atoms which subsequently undergo an electronic transition of energy \(3.62 \times 10^{-12} \mathrm{erg} .\) What color flame would this give rise to?

Short answer

The energy of the electronic transition is given as \(3.62 \times 10^{-12} \mathrm{erg}\), which we convert to Joules: \(E(J) = 3.62 \times 10^{-12} \mathrm{erg} \times 10^{-7} = 3.62 \times 10^{-19} \mathrm{J}\). We then calculate the wavelength using Planck's equation: \(\lambda = \frac{h \times c}{E} = \frac{6.626 \times 10^{-34} Js \times 3.00 \times 10^8 m/s}{3.62 \times 10^{-19} J} \approx 5.47 \times 10^{-7} m = 547 \mathrm{nm}\). This wavelength falls within the green region of the visible spectrum (495-570 nm), so the flame would appear green.

Step-by-step solution

Convert the energy to Joules

The energy given in the problem is in ergs, we need to convert it to Joules (J) to work with SI units. We know that 1 erg = \(10^{-7}\) Joules. So, we can convert the energy using the following relation: \[E (J) = E(erg) \times 10^{-7}\]

Calculate the wavelength of the emitted light

Now, we need to find the wavelength of the emitted light during the electronic transition. We can use Planck's equation to relate the energy and wavelength of the emitted light: \[E = h \times \frac{c}{\lambda}\] Here, E is the energy in Joules, h is Planck's constant (\(6.626 \times 10^{-34} Js\)), c is the speed of light (\(3.00 \times 10^8 m/s\)), and λ is the wavelength of the emitted light. We can rearrange the equation to find the wavelength: \[\lambda = \frac{h \times c}{E}\]

Compare the wavelength to the visible light spectrum

With the wavelength calculated, we can compare it to the visible light spectrum to find the color of the emitted light. The visible light spectrum has wavelengths ranging approximately from 400 nm (violet) to 700 nm (red). We can assign the colors as follows: - Violet: 400-450 nm - Blue: 450-495 nm - Green: 495-570 nm - Yellow: 570-590 nm - Orange: 590-620 nm - Red: 620-700 nm

Determine the color of the flame

Using the color ranges defined in the previous step, find the color of the flame corresponding to the calculated wavelength of the light emitted during the electronic transition of the barium atoms.

Key concepts

Barium

Barium is an alkaline earth metal with fascinating properties that make it recognizable during flame tests. This element, represented by the symbol Ba on the periodic table, is known for producing a distinctive green color in flame tests.
When a compound containing barium is introduced to a flame, it decomposes and releases barium atoms. These atoms are then excited, which means their electrons are moved to a higher energy level.
As these electrons return to their normal state, or ground state, they release energy. In the case of barium, this energy is predominantly emitted in the form of green light, providing a useful indicator for its presence.

• Barium flame tests are useful for identifying barium in a sample because of the unique color emitted during the test.
• Understanding barium's characteristic flame color helps in many chemical applications, including testing for the presence of barium in unknown substances.

Electronic Transition

Electronic transition is a fundamental concept in chemistry involving the movement of electrons between energy levels within an atom. When an electron absorbs energy, it moves from a lower energy level, or orbit, to a higher one—this is known as an excitation.
Each element has unique energy levels, so the amount of energy absorbed or released during an electron transition is unique to the that element.
In the flame test for barium, the electronic transition involves an energy change of approximately \(3.62 \times 10^{-12} \mathrm{erg}\), which is specific to barium. As the electrons fall back to their original energy levels, they release this energy in the form of light. This specific wavelength gives rise to the characteristic green color one observes in a flame test.

• Electronic transitions are central to understanding how different substances emit light of varying colors.
• These transitions are quantized, meaning only certain jump energies are possible, leading to the emission of specific colors.

Visible Light Spectrum

The visible light spectrum is the portion of the electromagnetic spectrum that can be detected by the human eye. It ranges approximately from wavelengths of 400 nm to 700 nm.
Each segment of this range corresponds to a different color, starting with violet at the lower wavelength end and red at the higher end.
Understanding this spectrum is crucial when working with flame tests, as it allows scientists to correlate observed colors with specific wavelengths and thereby identify substances based on their emitted light.

• Violet to blue light has shorter wavelengths and higher energy.
• Red to orange light has longer wavelengths and is lower in energy than blue and violet.

During a flame test, like that of barium, the specific wavelength of light emitted during an electron transition can be compared to the visible spectrum to determine its color—in this case, green. This allows for easy identification of elements based on observable flame color.