Question

The absorption spectrum of an atom consists of the wavelengths 200 nm, 300 nm, and 500 nm. To answer this question, first draw the atom's energy-level diagram and label the quantum states n=1,2,3,n=1,2,3, and so on. Part A What wavelengths are seen in the atom's emission spectrum? Express your answer in nanometers. λ₄ →1= Part B Express your answer in nanometers. λ₃ →1= Part C Express your answer in nanometers. λ₂ →1=

Answer 1

The wavelengths seen in the atom's emission spectrum would be the same as those in its absorption spectrum: 200 nm, 300 nm, and 500 nm, which correspond to the transitions from n=4 to n=1, n=3 to n=1, and n=2 to n=1, respectively.

Step-by-step explanation:

The absorption spectrum of an atom given the wavelengths 200 nm, 300 nm, and 500 nm can help us determine the atom's emission spectrum. According to the Bohr model of the atom, the absorption and emission spectra are essentially the inverse of one another. This means that the same wavelengths of light that are absorbed when an electron transitions from a lower to a higher energy level can be seen in the emission spectrum when the electron falls back to a lower energy level.

For instance, if an electron falls from the fourth energy level (n = 4) to the first energy level (n = 1), it would emit a photon corresponding to the same wavelength that would be absorbed to promote an electron from n = 1 to n = 4. Consequently, the emission spectrum would contain lines at the same wavelengths as those observed in the absorption spectrum.

Therefore, the answers to the various parts of your question are as follows:

For Part A, the wavelength λ4 → 1 is 200 nm.

For Part B, the wavelength λ3 → 1 is 300 nm.

For Part C, the wavelength λ2 → 1 is 500 nm.

Answer 2

To determine the wavelengths seen in the atom's emission spectrum, we need to consider transitions of electrons between energy levels. When an electron moves from a higher energy level (n ≥ 2) to a lower energy level (n = 1), it emits energy in the form of photons with specific wavelengths corresponding to the energy difference between the levels.

Here's how you can calculate these wavelengths step by step:

Step 1: Calculate the energy difference between energy levels n = 2, 3, and 4, and the ground state n = 1 using the formula:

ΔE = -R_H * (1/n_f^2 - 1/n_i^2)

Where ΔE is the energy difference, R_H is the Rydberg constant (approximately 2.18 × 10^(-18) J), and n_f and n_i are the final and initial quantum states, respectively.

For n = 2 to n = 1:

ΔE₂→₁ = -2.18 × 10^(-18) J * (1/1^2 - 1/2^2) = -1.22 × 10^(-18) J

For n = 3 to n = 1:

ΔE₃→₁ = -2.18 × 10^(-18) J * (1/1^2 - 1/3^2) = -1.94 × 10^(-19) J

For n = 4 to n = 1:

ΔE₄→₁ = -2.18 × 10^(-18) J * (1/1^2 - 1/4^2) = -1.36 × 10^(-19) J

Step 2: Use the energy-wavelength relationship to calculate the corresponding wavelengths for these energy differences:

E = h * c / λ

Where E is the energy, h is Planck's constant (approximately 6.626 × 10^(-34) J·s), c is the speed of light (approximately 3.00 × 10^8 m/s), and λ is the wavelength.

For ΔE₂→₁:

λ₂→₁ = h * c / |ΔE₂→₁|

λ₂→₁ = (6.626 × 10^(-34) J·s * 3.00 × 10^8 m/s) / (1.22 × 10^(-18) J) ≈ 1.63 × 10^(-7) m or 163 nm

For ΔE₃→₁:

λ₃→₁ = h * c / |ΔE₃→₁|

λ₃→₁ = (6.626 × 10^(-34) J·s * 3.00 × 10^8 m/s) / (1.94 × 10^(-19) J) ≈ 1.03 × 10^(-6) m or 1030 nm

For ΔE₄→₁:

λ₄→₁ = (6.626 × 10^(-34) J·s * 3.00 × 10^8 m/s) / (1.36 × 10^(-19) J) ≈ 1.46 × 10^(-6) m or 1460 nm

So, the wavelengths seen in the atom's emission spectrum are as follows:

Part A: λ₄→₁ ≈ 1460 nm

Part B: λ₃→₁ ≈ 1030 nm

Part C: λ₂→₁ ≈ 163 nm