To solve the problem, we need to follow these steps:
### Step 1: Determine the Ionization Energy
The ionization energy (IE) for a hydrogen atom in its ground state (n=1) is given by the formula:
\[ \text{IE} = \frac{13.6 \, \text{eV}}{n^2} \]
For hydrogen (Z=1) and in the ground state (n=1):
\[ \text{IE} = 13.6 \, \text{eV} \]
### Step 2: Calculate the Energy Absorbed by the Electron
According to the problem, the electron absorbs twice the ionization energy:
\[ \text{Energy absorbed} = 2 \times \text{IE} = 2 \times 13.6 \, \text{eV} = 27.2 \, \text{eV} \]
### Step 3: Calculate the Kinetic Energy of the Emitted Electron
The kinetic energy (KE) of the emitted electron after absorbing this energy is given by:
\[ \text{KE} = \text{Energy absorbed} - \text{IE} \]
Substituting the values:
\[ \text{KE} = 27.2 \, \text{eV} - 13.6 \, \text{eV} = 13.6 \, \text{eV} \]
### Step 4: Convert Kinetic Energy to Joules
To convert the kinetic energy from electron volts to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \text{KE} = 13.6 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]
\[ \text{KE} = 2.176 \times 10^{-18} \, \text{J} \]
### Step 5: Use De Broglie Wavelength Formula
The de Broglie wavelength (\( \lambda \)) of the emitted electron can be calculated using the formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot \text{KE}}} \]
Where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \) (mass of electron)
### Step 6: Substitute Values into the Formula
Substituting the values into the de Broglie wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 2.176 \times 10^{-18}}} \]
### Step 7: Calculate the Denominator
First, calculate the denominator:
\[ 2m \cdot \text{KE} = 2 \times 9.1 \times 10^{-31} \times 2.176 \times 10^{-18} \]
\[ = 3.96 \times 10^{-48} \]
Now take the square root:
\[ \sqrt{3.96 \times 10^{-48}} = 6.293 \times 10^{-24} \]
### Step 8: Calculate the Wavelength
Now substitute back into the wavelength formula:
\[ \lambda = \frac{6.626 \times 10^{-34}}{6.293 \times 10^{-24}} \]
\[ \lambda \approx 1.05 \times 10^{-10} \, \text{m} \]
### Final Answer
The wavelength of the emitted electron is approximately:
\[ \lambda \approx 3.34 \times 10^{-10} \, \text{m} \]
