Consider a hydrogen atom that goes from the n = 2 to the n = 1 state. (a) Calculate the wavelength, in nm, of the photon emitted. (b) What part of the electromagnetic spectrum is it?
To calculate the wavelength of the photon emitted when a hydrogen atom transitions from the n=2 to the n=1 state, we can use the equation:
1/λ = R * (1/n₁² - 1/n₂²)
Where:
- λ is the wavelength of the photon emitted
- R is the Rydberg constant (approximately 1.097 x 10^7 m⁻¹)
- n₁ is the initial state (n=2)
- n₂ is the final state (n=1)
Let's plug in the values:
1/λ = (1.097 x 10^7 m⁻¹) * (1/1² - 1/2²)
1/λ = (1.097 x 10^7 m⁻¹) * (1 - 1/4)
1/λ = (1.097 x 10^7 m⁻¹) * (3/4)
Multiplying these values:
1/λ = 8.2275 x 10^6 m⁻¹
Now, we can find the wavelength by taking the reciprocal:
λ = 1 / (8.2275 x 10^6 m⁻¹)
λ ≈ 1.214 x 10⁻⁷ m
To convert the wavelength to nm, we multiply by 10⁹:
λ (nm) = (1.214 x 10⁻⁷ m) * (10⁹ nm/m)
λ ≈ 121.4 nm
So, the wavelength of the photon emitted during the transition is approximately 121.4 nm.
To determine the part of the electromagnetic spectrum, we can refer to the commonly used ranges:
- Radio waves: >1 mm
- Microwaves: 1 mm - 1 μm
- Infrared: 1 μm - 700 nm
- Visible light: 400 - 700 nm
- Ultraviolet: 10 nm - 400 nm
- X-rays: <10 nm
- Gamma rays: <0.01 nm
Based on the calculated wavelength of approximately 121.4 nm, the emitted photon falls within the ultraviolet range.