What energy is required to excite a hydrogen atom in its n = 2 state to the n = 3 state ?
Isn't this the point of the Rydberg equation?
1/lambda= R(1/n^2 -1/m^2)
but E=plancksConstant*c/lambda
so E=plancksconstant*c * R (1/n^2-1/m^2)
c is the speed of light, R is the Rydberg constant, m,m are the two states.
What is neat about this method is it can be 'extended' to other atoms, by the multiplication Z^2 where Z is the atomic number.
To calculate the energy required to excite a hydrogen atom from the n = 2 state to the n = 3 state, we need to use the formula for the energy of an electron in a hydrogen atom:
E = -13.6 eV / n^2
where E is the energy, -13.6 eV is the ionization energy of hydrogen, and n is the principal quantum number.
First, let's find the energy for the n = 2 state:
E_initial = -13.6 eV / (2^2) = -13.6 eV / 4 = -3.4 eV
Next, we'll find the energy for the n = 3 state:
E_final = -13.6 eV / (3^2) = -13.6 eV / 9 = -1.51 eV
Finally, to find the energy required for excitation, we subtract the initial energy from the final energy:
∆E = E_final - E_initial = -1.51 eV - (-3.4 eV) = 1.89 eV
Therefore, the energy required to excite a hydrogen atom from the n = 2 state to the n = 3 state is 1.89 eV.