what is the wavelength of the photon emitted when an electron drops from a 4p orbital to a 2p orbital in a hydrogen atom
1/wavelength = R(1/2^2-1/4^2)
R = Rydberg constant = 1.097E7
To determine the wavelength of the photon emitted when an electron drops from a 4p orbital to a 2p orbital in a hydrogen atom, we need to use the equation:
λ = hc / ΔE
where:
λ is the wavelength of the photon,
h is the Planck's constant (h = 6.626 x 10^-34 J·s),
c is the speed of light (c = 3.00 x 10^8 m/s),
and ΔE is the energy difference between the initial and final states of the electron.
The energy difference (ΔE) can be obtained by subtracting the energy of the final state from the energy of the initial state. In this case, we need to use the formula for energy levels in hydrogen:
E = -13.6 eV / n^2
where E is the energy of the electron state, and n is the principal quantum number.
For the initial state, n = 4, so the energy of the 4p orbital is:
E initial = -13.6 eV / 4^2 = -13.6 eV / 16
For the final state, n = 2, so the energy of the 2p orbital is:
E final = -13.6 eV / 2^2 = -13.6 eV / 4
Now we can calculate the energy difference (ΔE) by subtracting the final state energy from the initial state energy:
ΔE = E initial - E final
Next, we can use this energy difference to calculate the wavelength (λ) using the equation given above:
λ = hc / ΔE
By substituting the values of h, c, and ΔE, we can determine the wavelength of the photon emitted when the electron drops from a 4p to a 2p orbital in a hydrogen atom.