What is the wavelength of the second photon emitted?

Question

A ground state H atom absorbs a photon of energy 2.093 x 10^-18 J, and its electron attains a higher energy level. As the atom relaxes, it emits two photons: one of wavelength 1282 nm to reach an intermediate level, and a second to reach the ground state. What is the wavelength of the second photon emitted?

Answer 1

The energy liberated by the electron falling back to the ground state must equal the energy absorbed, i.e. 2.093x10^-18 J. We can find the energy of the first photon given the wavelength of 1282 nm.

E = hc/λ where E = energy; h = Planck's constant; λ = wavelength

Energy = hc/λ = (6.626x10^-34 Js)(3x10⁸ m/s) / 1.282x10^-6 m

Energy = 1.551x10^-20 J which is the energy of the first photon

Energy of 2nd photon = 2.093x10^-18 J - 1.551x10^-20 J = 2.077x10^-20 J

Solving for λ we have λ = hc/E = (6.626x10^-34 Js)(3x10⁸ m/s) / 2.077x10^-20 J

λ = 9.571x10^-6 m = 9571 nm

Answer 2

Hello, Chris,

My strategy for this was subtract the energy of the 1282nm photon to leave an energy difference that would appear in the second emitted photon.

The energy of a photon comes from the Planck equation, E = hv, where v is the frequency in s^-1 (or Hertz), E is the energy in Joules, and h is Planck's constant of 6.6262x10^-34J*s. The relationship between wavelength and frequency for light is w*(v) = c, where w is the wavelength in nm, v is the frequency in s^-1, or Hertz, and c is the speed of light in m/sec (3.00x10⁸ m/s).

Combining Planck's and the speed of light equations allows us to calculate a photon's energy using the photon's wavelength: E = hc/w.

I had a table summarizing the steps, but it won't fit. Sorry. The energy of the 1280nm photon was 1.55E-19J. The difference between this and the absorbed energy will appear in the second emitted photon, which I found to be 1.938E-18J, which would have a wavelength of 103nm, which seems too low to me.

Bob