According to the quantum-mechanical model for the hydrogen atom, which of the following electron transitions would produce light with the longer wavelength:
2p --> 1s
or
3p --> 1s
I have the answer, which is 2p --> 1s, but can someone offer some help as to how this is determined?
Thanks for any help.
Make a diagram of the orbitals like so. You need to draw it out on a sheet of paper because I can't do it on the board.
-----------3p
-----------2p
------------1s
Now draw lines from 3p to 1s and from 2p to 1s. The most energy is emitted when the electron falls the most distance and the least energy is emitted when the electron falls the least distance.
The formula for energy is
E = hc/wavelength.
So if you want the longer wavelength, that means the denominator is larger which makes the right side of the equation lower and that means lower energy; therefore, you want to pick the transition that gives the least energy. I hope this helps.
To determine which electron transition would produce light with the longer wavelength, we can use the equation for calculating the wavelength of light emitted during a transition:
λ = c / ν
where λ is the wavelength, c is the speed of light, and ν is the frequency of the emitted light.
In the hydrogen atom, electron transitions are associated with the emission or absorption of photons of specific energy. The energy of a photon can be calculated using the formula:
E = -13.6 eV / n^2
where E is the energy, -13.6 eV is the ionization energy of hydrogen, and n is the principle quantum number of the electron shells involved in the transition.
The frequency of the emitted light can be related to the energy using Planck's equation:
E = h * ν
where E is the energy, h is Planck's constant (6.626 x 10^(-34) J·s), and ν is the frequency.
From these equations, we can conclude that the longer the wavelength, the lower the frequency and energy of the emitted light.
In the given electron transitions, 2p --> 1s and 3p --> 1s, we can compare the principle quantum numbers (n) involved. The transition 2p --> 1s involves a smaller difference in the principle quantum numbers (n = 2 to n = 1) compared to the transition 3p --> 1s (n = 3 to n = 1).
Since the energy difference decreases with decreasing principle quantum numbers, the transition 2p --> 1s will have a smaller energy and therefore a lower frequency and longer wavelength compared to the transition 3p --> 1s.
Thus, the electron transition 2p --> 1s would produce light with the longer wavelength.