Find the wavelength in nanometers of a photon emitted by a hydrogen atom decaying from the 3s state to the 2s state.
delta E = 2.18E-18 J(1/2^2 - 1/3^2), then
dE = hc/w. h is Planck's constant. c is speed of light in m/s. w is wavelength in m. Convert to nm. Post your work if you get stuck.
Thank you! It makes sense now.
To find the wavelength of the photon emitted by a hydrogen atom during the transition from the 3s state to the 2s state, we can use the Rydberg formula.
The Rydberg formula is given by:
1/λ = R_H * (1/n_f^2 - 1/n_i^2)
Here, λ represents the wavelength of the photon, R_H is the Rydberg constant for hydrogen (approximately 1.0974 × 10^7 m^-1), n_f is the principal quantum number of the final state, and n_i is the principal quantum number of the initial state.
In this case, the hydrogen atom is transitioning from the 3s state (n_i = 3) to the 2s state (n_f = 2). Therefore, we substitute these values into the formula:
1/λ = 1.0974 × 10^7 m^-1 * (1/2^2 - 1/3^2)
Calculating this further:
1/λ = 1.0974 × 10^7 m^-1 * (1/4 - 1/9)
1/λ = 1.0974 × 10^7 m^-1 * (9/36 - 4/36)
1/λ = 1.0974 × 10^7 m^-1 * (5/36)
1/λ = 0.152704 × 10^7 m^-1
Now, we can find the wavelength of the photon by taking the reciprocal of both sides:
λ = 1 / (0.152704 × 10^7 m^-1)
λ ≈ 6.55 × 10^-8 m
To convert this wavelength to nanometers, we multiply by 10^9 since 1 meter is equal to 10^9 nanometers:
λ = 6.55 × 10^-8 m * 10^9 nm/m
λ ≈ 655 nm
Therefore, the wavelength of the photon emitted by the hydrogen atom is approximately 655 nanometers.