The longest wavelength doublet absorption is observed at `589` and `589.6 nm`. Caiculate the frequency of each transition and energy difference between two excited states.

Options

A. `3.31xx10^(-22)kJ`
B. `3.31xx10^(-22)J`
C. `2.98xx10^(-21)J`
D. `3.0xx10^(-21)kJ`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert Wavelengths to Meters
We are given two wavelengths for the doublet absorption:
- \( \lambda_A = 589 \, \text{nm} \)
- \( \lambda_B = 589.6 \, \text{nm} \)

First, we convert these wavelengths from nanometers to meters:
\[ \lambda_A = 589 \, \text{nm} = 589 \times 10^{-9} \, \text{m} \]
\[ \lambda_B = 589.6 \, \text{nm} = 589.6 \times 10^{-9} \, \text{m} \]

### Step 2: Calculate Frequencies
We use the formula for frequency:
\[ \nu = \frac{c}{\lambda} \]
where \( c \) is the speed of light, approximately \( 3 \times 10^8 \, \text{m/s} \).

**For \( \nu_A \):**
\[ \nu_A = \frac{3 \times 10^8 \, \text{m/s}}{589 \times 10^{-9} \, \text{m}} = 5.093 \times 10^{14} \, \text{Hz} \]

**For \( \nu_B \):**
\[ \nu_B = \frac{3 \times 10^8 \, \text{m/s}}{589.6 \times 10^{-9} \, \text{m}} = 5.088 \times 10^{14} \, \text{Hz} \]

### Step 3: Calculate Energy Difference
The energy associated with a frequency is given by:
\[ E = h \nu \]
where \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \).

The energy difference \( \Delta E \) between the two states can be calculated as:
\[ \Delta E = E_A - E_B = h \nu_A - h \nu_B = h (\nu_A - \nu_B) \]

Substituting the values:
\[ \Delta E = 6.626 \times 10^{-34} \, \text{J s} \times (5.093 \times 10^{14} \, \text{Hz} - 5.088 \times 10^{14} \, \text{Hz}) \]
Calculating the difference in frequencies:
\[ \nu_A - \nu_B = 5.093 \times 10^{14} - 5.088 \times 10^{14} = 0.005 \times 10^{14} = 5 \times 10^{11} \, \text{Hz} \]

Now substituting back:
\[ \Delta E = 6.626 \times 10^{-34} \times 5 \times 10^{11} = 3.31 \times 10^{-22} \, \text{J} \]

### Final Answer
The energy difference between the two excited states is:
\[ \Delta E = 3.31 \times 10^{-22} \, \text{J} \]