To solve the problem, we need to determine the electronic states of the hydrogen atoms produced from the dissociation of the H₂ molecule and calculate the total kinetic energy (KE) of the dissociated hydrogen atoms. Here’s a step-by-step breakdown of the solution:
### Step 1: Understand the Dissociation Process
When the H₂ molecule absorbs a photon of wavelength 77.0 nm, it dissociates into two hydrogen atoms. The energy of the photon can be calculated using the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) 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 of the photon (77.0 nm = \( 77.0 \times 10^{-9} \, \text{m} \)).
### Step 2: Calculate the Energy of the Photon
Substituting the values into the equation:
\[
E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{77.0 \times 10^{-9} \, \text{m}}
\]
Calculating this gives:
\[
E \approx 2.58 \times 10^{-18} \, \text{J}
\]
### Step 3: Convert Energy to Electron Volts
To convert the energy from joules to electron volts (1 eV = \( 1.602 \times 10^{-19} \, \text{J} \)):
\[
E \approx \frac{2.58 \times 10^{-18}}{1.602 \times 10^{-19}} \approx 16.1 \, \text{eV}
\]
### Step 4: Determine the Electronic States of Hydrogen Atoms
The dissociation of H₂ produces two hydrogen atoms, which can be in different electronic states. The ground state of hydrogen has an energy of -13.6 eV. The possible electronic states for the hydrogen atoms after dissociation can be:
1. Both hydrogen atoms in the ground state (E = -27.2 eV).
2. One hydrogen atom in the ground state and the other in the first excited state (E = -13.6 eV for one and E = -3.4 eV for the other, total E = -17.0 eV).
3. Both hydrogen atoms in the first excited state (E = -3.4 eV each, total E = -6.8 eV).
### Step 5: Calculate Total Kinetic Energy (KE)
The total kinetic energy (KE) of the dissociated hydrogen atoms can be calculated by considering the energy of the photon absorbed and the total energy of the hydrogen atoms in the electronic states:
1. **Both in Ground State:**
\[
KE = E_{photon} - E_{total} = 16.1 \, \text{eV} - (-27.2 \, \text{eV}) = 43.3 \, \text{eV}
\]
2. **One in Ground State, One in First Excited State:**
\[
KE = 16.1 \, \text{eV} - (-17.0 \, \text{eV}) = 33.1 \, \text{eV}
\]
3. **Both in First Excited State:**
\[
KE = 16.1 \, \text{eV} - (-6.8 \, \text{eV}) = 22.9 \, \text{eV}
\]
### Summary of Results
- **Both in Ground State:** KE = 43.3 eV
- **One in Ground State, One in First Excited State:** KE = 33.1 eV
- **Both in First Excited State:** KE = 22.9 eV
