What was the energy of the original photon absorbed by the atom?

Question

A hydrogen atom absorbs a photon and its electron is excited to a higher energy level. Instead of immediately relaxing down to its original energy level, the electron first drops to an intermediate level, releasing a photon with a wavelength of 1282 nm. The electron then relaxes down to its original energy level, releasing a second photon with a wavelength of 102.6 nm. What was the energy of the original photon absorbed by the atom?

Answer 1

The important thing to remember here is the conservation of energy. In this case, this property can be simplified to "energy in = energy out."

We have three photons to deal with, which each have some energy value E. We have the original photon (E_i), the first photon released by the atom (E₁) and the second photon released by the atom (E₂). If we remember that "energy in = energy out" then we can write a new equation:

E_i = E₁ + E₂

This equation states that the energy of the initial photon is equal to the combined energy of the two photons released by the atom.

Now we need to determine values for E₁ and E₂. There are two relevant equations to understand:

1) E = hv

where E = energy of a specific photon in units of J (m*kg/s)

h = Planck's constant = 6.626*10^-34 m² kg / s

v = frequency of the photon in units of Hz (or s^-1)

2) c = λv

where c = speed of light constant = 3.00*10⁸ m/s

λ = wavelength of light in units of m

v = frequency of light in units of Hz (or s^-1)

Through the first equation, we see that we can calculate the energy of a photon if we have the frequency of the photon. However, we are not given the frequency of the two emitted photons, we are given their wavelengths. That's where we can use the second equation to find a relationship between wavelength and frequency, by dividing both sides of the equation by λ:

v = (c/λ)

Since we know both c and λ, we can replace v in equation 1.

E = h*(c/λ)

Going back to our original energy equation, we can substitute for the E₁ and E₂ values using the wavelengths of those photons, which we will call λ₁ and λ₂.

E_i = h*(c/λ₁) + h*(c/λ₂)

We have values for h and c (constants) as well as λ₁ and λ₂, so we are almost ready to solve for E_i. The last consideration we need to make is for units. You may have noticed that λ is typically in units of meters, but the wavelengths we are given are in nanometers. We need to convert these values before we continue.

1 m = 1*10⁹ nm

1282 nm * (1 m/1*10⁹ nm) = 1.282*10^-6 m

102.6 nm * (1 m/1*10⁹ nm) = 1.026*10^-7 m

Now we plug in all of our values:

E_i = (6.626*10^-34 m² kg / s) * (3.00*10⁸ m/s ÷ 1.282*10^-6 m) + (6.626*10^-34 m² kg / s) * (3.00*10⁸ m/s ÷ 1.026*10^-7 m)

E_i = 1.5505*10^-19 J + 19.3743*10^-19 J

E_i = 20.9248*10^-19 J

with significant figures,

E_i = 2.092*10^-18 J