Question

For a He*, calculate the wavelength of light (in m) that would be emitted for the orbital transition of n(initial) = 2 to n(final) = 1 The Rydberg constant is 1.09678

Answer 1

The wavelength of light that would be emitted during the orbital transition for a He* atom from n(initial) = 2 to n(final) = 1, using the Rydberg constant, is 6.835 × 10⁻¸ meters.

Step-by-step explanation:

The student is asking about the wavelength of light emitted from a He* (excited Helium) atom when an electron transitions from the second energy level to the first energy level. Using the Rydberg formula for the wavelengths of emitted radiation, we can calculate the wavelength by the formula:

λ = R × (1 / n²(final) - 1 / n²(initial))⁻¹

In this case, n(final) = 1 and n(initial) = 2, and given the Rydberg constant R is 1.09737 × 10⁷ m⁻¹, the formula simplifies to: λ = (1.09737 × 10⁷ m⁻¹) × (1 / 1² - 1 / 2²)⁻¹

To find λ, we perform the calculation:

λ = (1.09737 × 10⁷ m⁻¹) × (1 - 1/4)⁻¹

λ = (1.09737 × 10⁷ m⁻¹) × (3/4)⁻¹

λ = (1.09737 × 10⁷ m⁻¹) × 4/3

λ = 1.46316 × 10⁷ m⁻¹

Then the wavelength λ is: λ = 1 / 1.46316 × 10⁷ m⁻¹ = 6.835 × 10⁻¸ m

Therefore, the wavelength of light emitted for the transition from n = 2 to n = 1 in a He* atom is 6.835 × 10⁻¸ meters.