For sodium light, the two yellow lines occur at `lambda_(1) and lambda_(2)` wavelengths. If the mean of these two is `6000Å` and `|lambda_(2)-lambda_(1)|=6Å`, then the approximate energy difference between the two levels corresponding to `lambda_(1) and lambda_(2)` is

Options

A. `2xx10^(-3)eV`
B. `2eV`
C. `2000eV`
D. `2xx10^(-6)eV`

Answer

To solve the problem, we need to find the approximate energy difference between two energy levels corresponding to the wavelengths \( \lambda_1 \) and \( \lambda_2 \) of sodium light.

### Step-by-Step Solution:

1. **Understanding the Mean Wavelength**:
Given that the mean of the two wavelengths is \( 6000 \, \text{Å} \):
\[ \frac{\lambda_1 + \lambda_2}{2} = 6000 \, \text{Å} \]
This implies:
\[ \lambda_1 + \lambda_2 = 12000 \, \text{Å} \quad \text{(Equation 1)} \]

2. **Understanding the Difference in Wavelengths**:
We are also given that the difference between the two wavelengths is:
\[ |\lambda_2 - \lambda_1| = 6 \, \text{Å} \]
This gives us:
\[ \lambda_2 - \lambda_1 = 6 \, \text{Å} \quad \text{(Equation 2)} \]

3. **Solving the Equations**:
We can solve these two equations simultaneously. From Equation 2, we can express \( \lambda_2 \) in terms of \( \lambda_1 \):
\[ \lambda_2 = \lambda_1 + 6 \]
Substituting this into Equation 1:
\[ \lambda_1 + (\lambda_1 + 6) = 12000 \]
Simplifying this:
\[ 2\lambda_1 + 6 = 12000 \]
\[ 2\lambda_1 = 11994 \]
\[ \lambda_1 = 5997 \, \text{Å} \]
Now substituting back to find \( \lambda_2 \):
\[ \lambda_2 = 5997 + 6 = 6003 \, \text{Å} \]

4. **Calculating the Energy Difference**:
The energy corresponding to a wavelength \( \lambda \) is given by:
\[ E = \frac{hc}{\lambda} \]
Therefore, the energy difference \( \Delta E \) between the two wavelengths is:
\[ \Delta E = E_1 - E_2 = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} \]
Factoring out \( hc \):
\[ \Delta E = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) \]
This can be rewritten as:
\[ \Delta E = hc \cdot \frac{\lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \]

5. **Substituting Values**:
Now substituting \( \lambda_1 = 5997 \, \text{Å} \), \( \lambda_2 = 6003 \, \text{Å} \), and \( \lambda_2 - \lambda_1 = 6 \, \text{Å} \):
\[ \Delta E = hc \cdot \frac{6 \times 10^{-10}}{(5997 \times 10^{-10})(6003 \times 10^{-10})} \]
Using \( h = 6.63 \times 10^{-34} \, \text{J s} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \Delta E = (6.63 \times 10^{-34})(3 \times 10^8) \cdot \frac{6 \times 10^{-10}}{(5997 \times 6003) \times 10^{-20}} \]

6. **Calculating the Final Value**:
Calculate \( \Delta E \) and convert it to electron volts by dividing by \( 1.6 \times 10^{-19} \):
\[ \Delta E \approx 2 \times 10^{-3} \, \text{eV} \]

### Final Answer:
The approximate energy difference between the two levels corresponding to \( \lambda_1 \) and \( \lambda_2 \) is \( 2 \times 10^{-3} \, \text{eV} \).