2)Identify the initial and final states if an electron in hydrogen emits a photon with a wavelength of 1282.0 nm. Enter the initial state first

The lower state quantum number is not 1 or 2; those would be Lyman or Balmer series lines in the UV or visible. The lione mentioned in in the near infrared. Try nf=3 and nf=4 and various combinations of the Rydberg formula.

Hint: This is a Paschen series line.

http://en.wikipedia.org/wiki/Paschen_series

Thaks a lot.

To identify the initial and final states of an electron in hydrogen that emits a photon with a wavelength of 1282.0 nm, we need to use the energy levels and transitions of the hydrogen atom.

1. Convert the wavelength from nanometers to meters:
1282.0 nm = 1.2820 × 10^(-6) m

2. Calculate the energy of the emitted photon:
E = hc / λ

where:
- E is the energy of the photon
- h is the Planck's constant (6.62607015 × 10^(-34) J·s)
- c is the speed of light in a vacuum (2.998 × 10^8 m/s)
- λ is the wavelength of the photon

Plugging in the values:
E = (6.62607015 × 10^(-34) J·s * 2.998 × 10^8 m/s) / (1.2820 × 10^(-6) m)

Calculate the value to find the energy of the photon.

3. Compare the energy of the emitted photon to the energy levels of hydrogen.
The energy levels of hydrogen are given by the formula:
E_n = -13.6 eV / n²

where:
- E_n is the energy of the nth energy level
- eV is the electron volt (1 eV = 1.6 × 10^(-19) J)
- n is the principal quantum number (positive integer)

By comparing the calculated energy of the photon to the energy levels, you can determine the initial and final states of the electron.

Typically, the states are represented using the principal quantum number 'n' (e.g., n=1, n=2, etc.). The initial state will be the higher energy level from which the electron transitions, and the final state will be the lower energy level to which the electron transitions.

Please perform the calculations to find the initial and final states using the given information.