Ionization Energy of Hydrogen Atom:
To calculate the ionization energy of a hydrogen atom, we need to determine the energy required to remove an electron from the ground state.

The energy of an electron in the nth energy level of a hydrogen atom can be given by the formula:
E = -13.6/n^2 eV

Given that the Paschen series limit occurs at a wavelength of 8205.8 Angstrom, we can calculate the corresponding energy using the equation:
E = hc/λ

Where:
h = Planck's constant = 6.626 x 10^-34 Js
c = speed of light = 3 x 10^8 m/s
λ = wavelength in meters

Converting the wavelength to meters:
8205.8 Angstrom = 8205.8 x 10^-10 meters

Calculating the energy:
E = (6.626 x 10^-34 Js * 3 x 10^8 m/s) / (8205.8 x 10^-10 m)
E ≈ 2.42 x 10^-19 J

Converting the energy to electron volts (eV):
1 eV = 1.6 x 10^-19 J
Ionization energy = E / 1.6 x 10^-19
Ionization energy ≈ 1.51 eV

Therefore, the ionization energy of a hydrogen atom is approximately 1.51 electron volts.

Wavelength of the Photon that Removes the Electron in the Ground State:
The wavelength of the photon that would remove the electron in the ground state can be calculated using the energy difference between the ground state and the ionization energy.

The energy of the electron in the ground state (n=1) is given by:
E1 = -13.6 eV

The energy difference between the ground state and the ionization energy is:
ΔE = Ionization energy - E1
ΔE ≈ 1.51 eV - (-13.6 eV)
ΔE ≈ 15.11 eV

Using the equation:
E = hc/λ

Rearranging the equation to solve for wavelength:
λ = hc/E

Converting the energy to joules:
ΔE = 15.11 eV * 1.6 x 10^-19 J/eV

Calculating the wavelength:
λ = (6.626 x 10^-34 Js * 3 x 10^8 m/s) / (15.11 eV * 1.6 x 10^-19 J/eV)
λ ≈ 9.12 x 10^-8 m

Therefore, the wavelength of the photon that would remove the electron in the ground state of the hydrogen atom is approximately 9.12 x 10^-8 meters or 912 nm.