An electron begins in the n = 7 orbital of a hydrogen atom. It then emits a photon of wavelength λ and falls to a lower energy level. Before it has a chance to change levels again, it absorbs a 10 eV photon and is ejected from the atom with a kinetic energy of 6.6 eV. What is the wavelength, λ, of the first photon emitted?

Answer 1

First, let's find the initial energy of the electron in the n = 7 orbital. The energy levels in a hydrogen atom can be found using the formula:

E_n = -13.6 eV / n^2

where E_n is the energy of the electron in the nth orbital.

So for n = 7, the initial energy is:

E_7 = -13.6 eV / 7^2 ≈ -0.277 eV

Now let's find the energy of the electron after it absorbs the 10 eV photon and is ejected from the atom with a kinetic energy of 6.6 eV.

E_final = E_kinetic + 10 eV
E_final = 6.6 eV + 10 eV = 16.6 eV

In order for the electron to change energy levels and eventually be ejected, it must lose energy by emitting a photon with a certain wavelength. The energy of the emitted photon can be found using the change in energy of the electron:

E_photon = E_final - E_7
E_photon = 16.6 eV - (-0.277 eV) ≈ 16.877 eV

Now we can use the Planck-Einstein relation to find the wavelength of the emitted photon:

E_photon = (hc) / λ

where h is Planck's constant (4.136 x 10^-15 eV*s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength.

Rearranging the equation to find λ:

λ = (hc) / E_photon

Plugging in the values:

λ = (4.136 x 10^-15 eV*s * 3 x 10^8 m/s) / 16.877 eV ≈ 7.34 x 10^-8 m

So the wavelength of the emitted photon is approximately λ ≈ 734 nm, which falls in the near-infrared (NIR) region of the electromagnetic spectrum.