The electron in the n = 5 level of a hydrogen atom emits a photon with a wavelength of 1280 nm. To what energy level does the electron move?
Thank you, sir.
A hydrogen Atom makes a transition between two energy states and emits a photon with a wavelength of 434.2nm. What is the frequency? Does anyone know how to do this?
did you find that electron actually moved or did it stay in place?
3*10^8 / 4.342*10^-7 = frequency (units are seconds^-1)
When the electron in a hydrogen atom moves from n = 5 to n = 1, light is emitted.
Calculate the energy of the light?
To determine the energy level to which the electron moves, we can make use of the Rydberg formula, which relates the wavelengths of light emitted or absorbed by an electron in a hydrogen atom to the energy levels involved. The formula is given as:
1/λ = R * (1/nf² - 1/ni²)
Where:
- λ is the wavelength of the photon emitted or absorbed
- R is the Rydberg constant (approximately equal to 1.097 x 10^7 m⁻¹)
- nf is the final energy level
- ni is the initial energy level
In this case, we are given λ = 1280 nm. However, it is typically easier to work with wavelengths in meters, so we will convert the value to meters:
λ = 1280 nm = 1280 x 10^(-9) m
Now, we can rearrange the equation to solve for nf:
1/λ = R * (1/nf² - 1/ni²)
R * (1/nf² - 1/ni²) = 1/λ
1/nf² - 1/ni² = 1/(R * λ)
1/nf² = (1/(R * λ)) + 1/ni²
nf² = 1 / ((1/(R * λ)) + 1/ni²)
nf = √(1 / ((1/(R * λ)) + 1/ni²))
Plugging in the given values:
nf = √(1 / ((1/(1.097 x 10^7 m⁻¹ * 1280 x 10^(-9) m)) + 1/5²))
Now we can solve for nf.
1/wavelength = (2.180 x 10^-18/hc)x(1/n1^2-1/n2^2)
n1 = unknown
n2 = 5