To solve the problem, we will follow these steps:
### Step 1: Calculate the Ionization Energy of the Hydrogen Atom
The ionization energy of a hydrogen atom can be calculated using the formula derived from Bohr's model of the atom:
\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]
where:
- \( E_n \) is the energy of the electron at level \( n \),
- \( n \) is the principal quantum number.
For the ground state of hydrogen, \( n = 1 \):
\[
E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV}
\]
The ionization energy is the energy required to remove the electron completely from the atom, which means bringing the energy to zero. Therefore, the ionization energy \( E_{ion} \) is:
\[
E_{ion} = 0 - E_1 = 0 - (-13.6 \, \text{eV}) = 13.6 \, \text{eV}
\]
### Step 2: Calculate the Wavelength of the Photon Required to Ionize the Hydrogen Atom
To find the wavelength of the photon that would remove the electron from the ground state, we can use the relationship between energy and wavelength given by Planck's equation:
\[
E = \frac{hc}{\lambda}
\]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3.00 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength in meters.
First, we need to convert the ionization energy from electron volts to joules. The conversion factor is:
\[
1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}
\]
Thus, the ionization energy in joules is:
\[
E_{ion} = 13.6 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 2.179 \times 10^{-18} \, \text{J}
\]
Now, substituting the values into Planck's equation:
\[
\lambda = \frac{hc}{E}
\]
Substituting \( h \), \( c \), and \( E_{ion} \):
\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3.00 \times 10^8 \, \text{m/s})}{2.179 \times 10^{-18} \, \text{J}}
\]
Calculating this gives:
\[
\lambda \approx 9.12 \times 10^{-7} \, \text{m} = 912 \, \text{nm} = 9120 \, \text{Å}
\]
### Final Answers:
a. The ionization energy of the hydrogen atom is **13.6 eV**.
b. The wavelength of the photon that would remove the electron in the ground state of the hydrogen atom is **912 Å**.
