Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from n = 6 to n = 1.
Where the hell did you get 2.18x10^-18 from
@drbob222 shes right thoooo
To calculate the energy of a photon emitted during an electronic transition in a hydrogen atom, you can use the Rydberg formula:
1/λ = R_H * (1/n_f^2 - 1/n_i^2)
Where:
- λ is the wavelength of the photon emitted.
- R_H is the Rydberg constant for hydrogen, approximately 1.097 × 10^7 m^-1.
- n_f is the final energy level of the electron.
- n_i is the initial energy level of the electron.
In this case, the electron undergoes a transition from n = 6 (initial level) to n = 1 (final level).
Let's plug in the values:
1/λ = R_H * (1/1^2 - 1/6^2)
1/λ = R_H * (1 - 1/36)
1/λ = R_H * (35/36)
Now, solve for λ:
λ = 36/(35 * R_H)
λ = 36/(35 * 1.097 × 10^7 m^-1)
λ ≈ 8.228 × 10^-8 meters
Now that we have the wavelength, we can use the energy-wavelength relationship for photons:
E = hc/λ
Where:
- E is the energy of the photon.
- h is the Planck constant, approximately 6.626 × 10^-34 J·s.
- c is the speed of light in a vacuum, approximately 3.00 × 10^8 m/s.
Plug in the values:
E = (6.626 × 10^-34 J·s * 3.00 × 10^8 m/s) / (8.228 × 10^-8 meters)
E ≈ 2.411 × 10^-18 J
Therefore, the energy of the photon emitted during the transition from n = 6 to n = 1 in a hydrogen atom is approximately 2.411 × 10^-18 Joules.
for both? and then add them?
No. E = 2.18E-18J x (1/1 - 1/36) =
E = 2.18E-18J x (1-0.0278) = ?
I just wrote the formula and substituted so you could see where the numbers came from.
n1 = 1 and 1^2 = 1
n2 = 6 and 6^2 = 36
E = 2.18E-18J x (1/n1^2 - 1/n2^2)
E = 2.18E-18 J x (1/1 - 1/36)
Solve for E.