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?
You have two problems in one here. The first one is to start with the electron in its ground state and determine to which level (shell) the photon takes it. The second part starts with the electron in that as a starting position gives the wavelength and asks for the final shell.
#1.
delta E = hc/wavelength
Substitute h, c & wavelength and calculate delta E. Then use
delta E = 2.180E18 J x (1/1 - 1/n^2)
The 1/1 is 1/n^2 for the electron in n = 1. The n^2 is the shell to which the electron is being promoted. Substitute and solve for n.
Second part.
delta E = hc/wavelength. Substitute and solve for delta E.
delta E = 2.180E18J x (1/x^2 - 1/n^2)
The n^2 is the shell to which the electron has been promoted from the previous part and the x = the final destination of the electron. Substitute and solve for x, then
ok, i think i got 5 for an answer
do you think this is correct??
To determine the final state of the hydrogen atom, we can analyze the energy changes involved.
First, let's find the initial energy of the ground state hydrogen atom. We can use 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 principal quantum number. For the ground state, n = 1, so the initial energy is -13.6 eV.
Next, let's find the energy of the absorbed photon of light. We can use the formula:
E = h * c / λ
where E is the energy, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the light. Converting the wavelength of 94.96 nm to meters, we get 9.496 x 10^-8 m. Plugging these values into the formula, we find that the absorbed energy is approximately 2.081 x 10^-18 J.
Now, let's find the final energy of the hydrogen atom after it emits a photon of 95 nm. Using the same formula, we find that the energy of this photon is approximately 2.086 x 10^-18 J.
Since the hydrogen atom absorbed and emitted photons with very similar energies, it suggests that the initial and final states of the atom are very close in energy. In this case, the final state of the hydrogen atom is likely to be the first excited state, which has a principal quantum number of n = 2.
Therefore, the final state of the hydrogen atom is the first excited state with a principal quantum number of n = 2.