A free electron is captured by a proton forming a Hydrogen atom. If the electron ends up in the ground state then what is the wavelength of the photon it emits during this process?

I don't think professor clark would be happy to know you're not doing your own hw!

The energy of the photon is the binding energy of the hydrogen ground state, 13.6 eV = 2.18*10^-18 J

Set that equal to h c/(wavelength) and solve for the wavelength. It will be in the far "vacuum" ultraviolet.

wavelength = h c/2.18*10^-18 J

h is Planck's constant and c is the speed of light

To determine the wavelength of the photon emitted during the transition of an electron from a higher energy state to the ground state in a hydrogen atom, we can use the formula for calculating the wavelength of electromagnetic radiation:

λ = hc/E

Where:
λ = wavelength of the photon
h = Planck's constant (6.626 × 10^-34 J·s)
c = speed of light in a vacuum (2.998 × 10^8 m/s)
E = energy of the transition

The energy of the transition can be determined using the Rydberg formula:

E = RH(1/n_initial^2 - 1/n_final^2)

Where:
RH = Rydberg constant (2.18 × 10^-18 J)
n_initial = initial energy level
n_final = final energy level

In this case, since the electron is transitioning from a higher energy state to the ground state, we can set n_initial as 2 (corresponding to the second energy level) and n_final as 1 (corresponding to the ground state).

Plugging in these values, we have:

E = RH(1/2^2 - 1/1^2)
E = RH(1/4 - 1/1)
E = RH(3/4)

Now we can calculate the wavelength of the emitted photon by substituting the value of E into the first equation:

λ = hc/E
λ = (6.626 × 10^-34 J·s) × (2.998 × 10^8 m/s) / [RH(3/4)]

Finally, we substitute the value of the Rydberg constant, RH:

λ = (6.626 × 10^-34 J·s) × (2.998 × 10^8 m/s) / [(2.18 × 10^-18 J)(3/4)]

Evaluating this expression will give us the wavelength of the photon emitted during the process of an electron transitioning from a higher energy state to the ground state in a hydrogen atom.