H2P4: PHOTON EMISSION
An electron in Li2+ falls from n = 2 to the ground state. Calculate the wavenumber, (νˉ), of the emitted photon.
1/wavelength = wave number = 9*1.09737E7(1/1 - 1/4) =
That comes from
1/wavelength = RZ^2(1/n1^2 - 1/n2^2)
n1 = 1 and 1^2 = 1
n2 = 2 and 2^2 = 4
R = 1.09737E7
Z = atomic number = 3^2 = 9
To calculate the wavenumber (νˉ) of the emitted photon, we can use the equation:
νˉ = R * (1/ni^2 - 1/nf^2)
Where:
- νˉ is the wavenumber of the emitted photon
- R is the Rydberg constant (1.0973731568508 x 10^7 m⁻¹)
- ni is the initial energy level (2 in this case)
- nf is the final energy level (1 for the ground state)
Now let's substitute the values into the equation:
νˉ = (1.0973731568508 x 10^7 m⁻¹) * (1/2^2 - 1/1^2)
Simplifying the equation further:
νˉ = (1.0973731568508 x 10^7 m⁻¹) * (1/4 - 1/1)
νˉ = (1.0973731568508 x 10^7 m⁻¹) * (1/4 - 1)
Now evaluate the expression:
νˉ = (1.0973731568508 x 10^7 m⁻¹) * (1/4 - 4/4)
νˉ = (1.0973731568508 x 10^7 m⁻¹) * (-3/4)
Calculating:
νˉ = -8.2307979278816 x 10^6 m⁻¹
Therefore, the wavenumber (νˉ) of the emitted photon is approximately -8.2307979278816 x 10^6 m⁻¹.