What is the energy of the original photon?

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 1324 nm. The electron then relaxes down to its original energy level, releasing a second photon with a wavelength of 126.9 nm. What was the energy of the original photon absorbed by the atom?

Answer 1

Hello, Chris,

Use Plank's equation E = hv, where E is energy (in Joules), h is Plank's constant (Js), and v is the frequency of the light (1/s or s^-1). [s is seconds]. Remember that 1nm = 1E-9 meter. We need to convert the wavelengths to frequencies by using the relation v = c/ʎ, where lambda is the wavelength and v is frequency. We can also substitute the second expression into the first to get E = hc/ʎ, which is what I used. h = Js

Bob