A ground state hydrogen atom absorbs a photon of light having a wavelength of 94.96 nm. It then gives off a photon having a wavelength of 95 nm. What is the final state of the hydrogen atom?
I think I got 5 for an answer, but I am not sure if I did this right or not
I ran through the calc quickly (and could have made an error) but I came up with 5 for where the electron was promoted to (from ground state of n = 1 to n = 5) but I did not get 5 for the final resting place. I will look for the error if you want to post your work.
well i got 5 for the first part but I thought that the delta E was the same for both because the wavelengths were pretty much the same, so i got 1 as the final but that's only because that made both of the equations exactly the same
You are absolutely right. And I should have notice that too but didn't until just a second before reading your latest post. The first photon sends it from n = 1 to n = 5 and the emission of energy then sends it back to n = 1. The first part of your observation is also right. Since the first photon and the second photon are ALMOST the same, that MUST be true because it emits as much energy on the way down as it took to get it up there (to n = 5) in the first place. :-).
I'm not sure what you mean by "made both equations exactly the same."
The second equation is
E = 6.626E-34*3E8/95E-9 = 2.0924E-18, then
2.09824E-18 = 2.180E-18*(1/x^2 - 1/25) which isn't the same as the first equation you solved.
x^2 = 1.00017E0 so x = 1.000
Ok I thought that before I submitted the question but I thought that seemed too easy! That you so much for your help though :)
To determine the final state of the hydrogen atom, we need to consider the energy changes that occur when a photon is absorbed and emitted. The energy of a photon is given by the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the photon.
First, let's calculate the energy of the absorbed photon using the given wavelength of 94.96 nm:
E₁ = (6.626 x 10^-34 J·s) * (3.00 x 10^8 m/s) / (94.96 x 10^-9 m)
Simplifying the calculation, we get:
E₁ = 2.084 x 10^-18 J
The photon is absorbed by the hydrogen atom, causing an electron to move to a higher energy level. The hydrogen atom's energy level can be represented by the principal quantum number, n. As the electron moves to a higher energy level, the value of n increases.
Next, we need to determine the energy of the emitted photon using the given wavelength of 95 nm:
E₂ = (6.626 x 10^-34 J·s) * (3.00 x 10^8 m/s) / (95 x 10^-9 m)
Again, simplifying the calculation, we get:
E₂ = 2.079 x 10^-18 J
The electron in the hydrogen atom then transitions back to a lower energy level, emitting a photon.
To find the final state of the hydrogen atom, we need to compare the energy of the absorbed and emitted photons.
From the calculations above, we can see that the energy of the absorbed photon (E₁ = 2.084 x 10^-18 J) is slightly higher than the energy of the emitted photon (E₂ = 2.079 x 10^-18 J). This indicates that the difference in energy between these two states is very small.
The principal quantum number, n, indicates the energy level of the electron in a hydrogen atom. For hydrogen, the energy of an electron at a given level is given by the equation E = -13.6 eV / n², where E is the energy in electron volts (eV) and n is the principal quantum number.
As the difference in energy between the absorbed and emitted photons is very small, we can assume that the electron transition occurs between two states with similar principal quantum numbers. By comparing the energy levels of hydrogen atoms, we find that the closest energy levels are n = 5 and n = 6.
Therefore, the final state of the hydrogen atom is likely when the electron transitions from the 6th energy level (n = 6) to the 5th energy level (n = 5).