A ground state hydrogen atom absorbs a photon of light having a wavelength of 93.03 nm. It then gives off a photon having a wavelength of 396.9 nm. What is the final state of the hydrogen atom? Values for physical constants can be found here.
Now I just did one using the same relationships between frequency, period, speed of light, wavelength and energy. This one is the same.
http://www.jiskha.com/display.cgi?id=1455410600
note,
absorbed high energy
radiated much lower energy
therefore, energy of atom went up.
To determine the final state of the hydrogen atom, we need to understand the energy levels associated with it. According to the Bohr model, the energy levels of a hydrogen atom are given by the equation:
E = -13.6 eV / n^2
where E is the energy, n is the principal quantum number, and -13.6 eV is the ionization energy of hydrogen.
When a hydrogen atom absorbs a photon, it transitions from a lower energy level to a higher energy level. The energy change associated with this transition is given by the equation:
ΔE = E_final - E_initial
where ΔE is the energy change, E_final is the energy of the higher energy level, and E_initial is the energy of the lower energy level.
To find the initial energy level, we can use the formula for the energy of a photon:
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.
Given that the wavelength of the absorbed photon is 93.03 nm, we can calculate its energy:
E_initial = hc / λ_initial
= (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s) / (93.03 x 10^-9 m)
= 2.240 x 10^-18 J
To find the final energy level, we can use the same formula with the wavelength of the emitted photon:
E_final = hc / λ_final
= (6.626 x 10^-34 J·s)(3.00 x 10^8 m/s) / (396.9 x 10^-9 m)
= 4.977 x 10^-19 J
Now we can calculate the energy change:
ΔE = E_final - E_initial
= (4.977 x 10^-19 J) - (2.240 x 10^-18 J)
= -1.742 x 10^-18 J
The negative sign indicates that the hydrogen atom lost energy during this transition. To determine the final state, we need to find the energy level that corresponds to this energy change.
Using the equation for the energy levels:
ΔE = -13.6 eV / (n_final^2) - (-13.6 eV / (n_initial^2))
We can rearrange the equation to solve for n_final:
n_final^2 = (13.6 eV * ΔE) / (-13.6 eV) + n_initial^2
Since n_initial is equal to 1 for the ground state, we have:
n_final^2 = (13.6 eV * ΔE) / (-13.6 eV) + 1^2
= ΔE / (-1.0 eV) + 1
= (-1.742 x 10^-18 J) / (-1.6 x 10^-19 J) + 1
= 10.888 + 1
= 11.888
Taking the square root, we find:
n_final = √11.888
n_final ≈ 3.450
Therefore, 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 396.9 nm corresponds to the n = 3 energy level.