The energy difference between two electronic states is 399.1 kJ `mol^(-1)`. Calculate the wavelength and frequency of light emitted when an electron drops from a higher to a lower state. (Planck's constant, `h = 3.98 xx 10^(-13) kJ mol^(-1))`.

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Understand the relationship between energy, frequency, and wavelength
The energy difference (ΔE) between two electronic states is related to the frequency (ν) of the emitted light by the equation:
\[ \Delta E = h \cdot \nu \]
where:
- ΔE is the energy difference (in kJ/mol),
- h is Planck's constant (in kJ/mol·s),
- ν is the frequency (in Hz).

### Step 2: Rearranging the equation to find frequency
From the equation above, we can express frequency as:
\[ \nu = \frac{\Delta E}{h} \]

### Step 3: Substitute the values
Given:
- ΔE = 399.1 kJ/mol
- h = 3.98 × 10^(-13) kJ/mol·s

Substituting these values into the equation:
\[ \nu = \frac{399.1 \, \text{kJ/mol}}{3.98 \times 10^{-13} \, \text{kJ/mol·s}} \]

### Step 4: Calculate the frequency
Now, perform the calculation:
\[ \nu = \frac{399.1}{3.98 \times 10^{-13}} \approx 1.003 \times 10^{15} \, \text{Hz} \]
For simplicity, we can round this to:
\[ \nu \approx 1.0 \times 10^{15} \, \text{Hz} \]

### Step 5: Use the frequency to find the wavelength
The relationship between the speed of light (c), wavelength (λ), and frequency (ν) is given by:
\[ c = \lambda \cdot \nu \]
Rearranging this to find wavelength gives:
\[ \lambda = \frac{c}{\nu} \]

### Step 6: Substitute the values for speed of light and frequency
The speed of light (c) is approximately:
\[ c = 3.0 \times 10^8 \, \text{m/s} \]
Substituting the values:
\[ \lambda = \frac{3.0 \times 10^8 \, \text{m/s}}{1.0 \times 10^{15} \, \text{Hz}} \]

### Step 7: Calculate the wavelength
Now, perform the calculation:
\[ \lambda = 3.0 \times 10^{-7} \, \text{m} \]
To convert this to nanometers (1 m = 10^9 nm):
\[ \lambda = 3.0 \times 10^{-7} \, \text{m} = 300 \, \text{nm} \]

### Final Results
- Frequency (ν) ≈ \(1.0 \times 10^{15}\) Hz
- Wavelength (λ) ≈ 300 nm