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