Background: A monochromatic laser is exciting hydrogen atoms from the n=2 to state to the n=5 state.
Question: What is the shortest wavelength observed?
-I believe that it will be from n=5 to n=4, but I am not sure what equation to use or how to use the n values...
Answer: ?? nm
short wavelenght, highest energy. Highest energy means greatest possible change.
So calculate yours, then try 2 to 5
To determine the shortest wavelength observed when a monochromatic laser excites hydrogen atoms from the n=2 state to the n=5 state, we need to use the equation for the energy of an electron in a hydrogen atom.
The equation for the energy of an electron in a hydrogen atom is given by:
E = -13.6 eV / n^2
Where:
E is the energy of the electron in electron volts (eV).
n is the principal quantum number that represents the energy level of the electron.
For the transition from the n=5 state to the n=4 state, we can calculate the energy difference using:
ΔE = E_initial - E_final
Substituting the respective values, we have:
E_initial = -13.6 eV / (5^2)
E_final = -13.6 eV / (4^2)
Calculating the energy difference:
ΔE = (-13.6 eV / 25) - (-13.6 eV / 16)
To convert this energy difference to the corresponding wavelength, we can use the equation:
Energy = Planck's constant * speed of light / wavelength
Rearranging the equation to solve for wavelength:
Wavelength = Planck's constant * speed of light / Energy
By substituting the energy difference into the equation, we can calculate the shortest observed wavelength.