An electron of a hydrogen atom falling from n=3 in the excited state to back to ground state gives off about 12eV. what form the "light" will it give off?
I would do this.
1/wavelength = R(1/1 - 1/9)
R you can find in tables in your book or on the web. Calculate wavelength and from the wavelength look in your book or on the web for the spectrum and see where that wavelength falls.
To determine the form of light emitted when an electron of a hydrogen atom falls from an excited state to the ground state, we need to understand the energy levels in a hydrogen atom.
The energy levels in hydrogen atoms are quantized and can be described by the Rydberg formula:
1/λ = R * (1/n₁² - 1/n₂²)
where λ is the wavelength of light emitted, R is the Rydberg constant (approximately 1.097 × 10⁷ m⁻¹), and n₁ and n₂ are the principal quantum numbers of the initial and final energy states, respectively.
In this case, the electron is falling from n=3 (initial state) to the ground state (n=1). We can calculate the wavelength of the emitted light using the Rydberg formula:
1/λ = R * (1/n₁² - 1/n₂²)
1/λ = 1.097 × 10⁷ m⁻¹ * (1/3² - 1/1²)
1/λ = 1.097 × 10⁷ m⁻¹ * (1/9 - 1/1)
1/λ = 1.097 × 10⁷ m⁻¹ * (1/9 - 1)
1/λ = 1.097 × 10⁷ m⁻¹ * (-8/9)
1/λ = -9.763 × 10⁶ m⁻¹
To find the wavelength (λ), we can take the reciprocal of both sides:
λ = -1 / (-9.763 × 10⁶ m⁻¹)
λ = 1.025 × 10⁻⁷ m (approximately)
The wavelength calculated is approximately 1.025 × 10⁻⁷ meters. This wavelength corresponds to electromagnetic radiation in the ultraviolet region of the spectrum.
Therefore, when the electron of a hydrogen atom falls from the n=3 excited state to the ground state, it emits ultraviolet light.