To calculate the frequency of infrared photons that can be absorbed by the HD molecule, we will follow these steps:
### Step 1: Understand the Energy Change
The energy change (ΔE) associated with the transition of the HD molecule can be calculated using the formula:
\[
\Delta E = E_1 - E_0
\]
where \( E_1 \) is the energy of the excited state and \( E_0 \) is the energy of the ground state.
### Step 2: Calculate the Ground State Energy
The total energy of the hydrogen molecule \( H_2 \) in its ground state is given as:
\[
E_0 = -31.675 \, \text{eV}
\]
### Step 3: Calculate the Excited State Energy
For the HD molecule, we can use the formula for the energy levels of the hydrogen atom:
\[
E_n = -\frac{R_H}{n^2}
\]
where \( R_H = 13.5984 \, \text{eV} \). For the first excited state (n=2):
\[
E_1 = -\frac{13.5984}{2^2} = -\frac{13.5984}{4} = -3.3996 \, \text{eV}
\]
### Step 4: Calculate the Energy Difference
Now, we can calculate the energy difference (ΔE):
\[
\Delta E = E_1 - E_0 = -3.3996 - (-31.675)
\]
\[
\Delta E = -3.3996 + 31.675 = 28.2754 \, \text{eV}
\]
### Step 5: Convert Energy to Joules
Next, we need to convert this energy from electron volts to joules. Using the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \):
\[
\Delta E = 28.2754 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV}
\]
\[
\Delta E \approx 4.53 \times 10^{-18} \, \text{J}
\]
### Step 6: Calculate Frequency
Using the formula \( E = h \nu \), where \( h = 6.6261 \times 10^{-34} \, \text{J s} \), we can solve for frequency \( \nu \):
\[
\nu = \frac{\Delta E}{h} = \frac{4.53 \times 10^{-18} \, \text{J}}{6.6261 \times 10^{-34} \, \text{J s}}
\]
\[
\nu \approx 6.83 \times 10^{15} \, \text{s}^{-1}
\]
### Final Answer
The frequency of infrared photons that can be absorbed by the HD molecule is approximately:
\[
\nu \approx 6.83 \times 10^{15} \, \text{s}^{-1}
\]
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