To find the longest wavelength of radiation required to remove the electron from Bohr's first orbit, we can follow these steps:
### Step 1: Understand the Ionization Energy
The ionization energy of the hydrogen atom in the ground state is given as 13.6 eV. This is the energy required to completely remove the electron from the first orbit.
### Step 2: Convert Ionization Energy to Joules
We need to convert the ionization energy from electron volts (eV) to joules (J) because we will use the Planck's constant in joules. The conversion factor is:
1 eV = 1.6 × 10^-19 J.
So, we calculate:
\[
E = 13.6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 2.176 \times 10^{-18} \, \text{J}.
\]
### Step 3: Use the Energy-Wavelength Relationship
The energy of a photon is related to its wavelength by the equation:
\[
E = \frac{hc}{\lambda}
\]
where:
- \( E \) is the energy,
- \( h \) 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.
### Step 4: Rearranging the Equation
We can rearrange the equation to solve for the wavelength \( \lambda \):
\[
\lambda = \frac{hc}{E}.
\]
### Step 5: Substitute Known Values
Now we substitute the known values into the equation:
\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{2.176 \times 10^{-18} \, \text{J}}.
\]
### Step 6: Calculate the Wavelength
Calculating the above expression:
\[
\lambda = \frac{1.9878 \times 10^{-25} \, \text{J m}}{2.176 \times 10^{-18} \, \text{J}} \approx 9.12 \times 10^{-8} \, \text{m}.
\]
### Step 7: Convert Wavelength to Angstroms
To convert meters to angstroms (1 angstrom = \( 10^{-10} \, \text{m} \)):
\[
\lambda = 9.12 \times 10^{-8} \, \text{m} \times \frac{1 \, \text{Å}}{10^{-10} \, \text{m}} = 912 \, \text{Å}.
\]
### Final Answer
The longest wavelength of the radiation required to remove the electron from Bohr's first orbit is approximately **912 Å**.
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