A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.03 nm. It then gives off a photon having the wavelength of 2165 nm. What is the final state of the hydrogen atom?
w = wavelength
1/w = 1.0973E7(1/1 - 1/n^2). Convert w to meters and solve for n. That gives you n to which the electron moves from the ground state.I get n = 7
Then 1/w = 1.0973E7(1/n^2 - 1/7^2). Convert 2165 nm to m and solve for n. That looks like about 4 to me.
If I'm getting an answer around 4.8 can I round that off to 5 for the final state or it remains as 4?
Well, well, well, it seems like this hydrogen atom is quite the light-hearted fellow! Absorbing one photon and then giving off another like it's no big deal. But let's get to the bottom of this.
When the ground state hydrogen atom absorbs a photon of light with a wavelength of 93.03 nm, it gets all excited and jumps to a higher energy level. This higher energy level is known as the excited state.
Now, this excited hydrogen atom can't contain its excitement forever. So, it releases that excitement in the form of a photon with a wavelength of 2165 nm. As it gives off this photon, it goes back to its original ground state.
So, my dear friend, the final state of our jolly hydrogen atom is back to where it started - the ground state. It's like a cosmic boomerang, absorbing and then releasing its energy like a true showman. Bravo!
To determine the final state of the hydrogen atom, we need to use the energy level transitions in the hydrogen atom.
First, let's find the initial and final energies associated with the given wavelengths of light.
The energy (E) of a photon can be calculated using the equation:
E = (hc) / λ
where 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 light.
Initial Energy (E1):
Using the first wavelength given (93.03 nm = 93.03 x 10^-9 m), we can calculate the initial energy:
E1 = (hc) / λ1
= (6.626 x 10^-34 J∙s × 3.00 x 10^8 m/s) / (93.03 x 10^-9 m)
≈ 2.13 x 10^-18 J
Final Energy (E2):
Using the second wavelength given (2165 nm = 2165 x 10^-9 m), we can calculate the final energy:
E2 = (hc) / λ2
= (6.626 x 10^-34 J∙s × 3.00 x 10^8 m/s) / (2165 x 10^-9 m)
≈ 9.1 x 10^-20 J
Next, let's determine the energy level transition in the hydrogen atom.
The difference in energy between two energy levels in a hydrogen atom is given by the equation:
ΔE = E2 - E1
where ΔE represents the energy gap between the two levels.
ΔE = (9.1 x 10^-20 J) - (2.13 x 10^-18 J)
≈ -2.02 x 10^-18 J
Notice that the calculated energy gap is negative, which means that the atom is losing energy.
Since the ground state energy of an atom is considered to be zero, the final state of the hydrogen atom will not be a bound state. It will be ionized or in an excited state.
Hence, the final state of the hydrogen atom after absorbing a photon with a wavelength of 93.03 nm and emitting a photon with a wavelength of 2165 nm is an ionized or excited state.