To solve the problem, we will follow these steps:
### Step 1: Understand the given information
We are given:
- The mean wavelength of sodium light, \( \lambda_{mean} = 6000 \, \text{Å} \)
- The difference between the two wavelengths, \( |\lambda_2 - \lambda_1| = 6 \, \text{Å} \)
### Step 2: Set up equations
From the mean wavelength, we can write:
\[
\frac{\lambda_1 + \lambda_2}{2} = 6000 \, \text{Å}
\]
This leads to:
\[
\lambda_1 + \lambda_2 = 12000 \, \text{Å} \quad \text{(Equation 1)}
\]
From the difference in wavelengths, we can write:
\[
\lambda_2 - \lambda_1 = 6 \, \text{Å} \quad \text{(Equation 2)}
\]
### Step 3: Solve the equations
Now we will 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 expression into Equation 1:
\[
\lambda_1 + (\lambda_1 + 6) = 12000
\]
\[
2\lambda_1 + 6 = 12000
\]
\[
2\lambda_1 = 12000 - 6
\]
\[
2\lambda_1 = 11994
\]
\[
\lambda_1 = 5997 \, \text{Å}
\]
Now substituting \( \lambda_1 \) back into the equation for \( \lambda_2 \):
\[
\lambda_2 = 5997 + 6 = 6003 \, \text{Å}
\]
### Step 4: Calculate the energy difference
The energy associated with a wavelength is given by the formula:
\[
E = \frac{12400}{\lambda} \quad \text{(in eV, where } \lambda \text{ is in Å)}
\]
Now, we need to find the energy difference \( E_2 - E_1 \):
\[
E_2 - E_1 = \frac{12400}{\lambda_2} - \frac{12400}{\lambda_1}
\]
\[
= 12400 \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} \right)
\]
Substituting \( \lambda_1 = 5997 \, \text{Å} \) and \( \lambda_2 = 6003 \, \text{Å} \):
\[
E_2 - E_1 = 12400 \left( \frac{1}{6003} - \frac{1}{5997} \right)
\]
Calculating the difference:
\[
\frac{1}{6003} - \frac{1}{5997} = \frac{5997 - 6003}{6003 \times 5997} = \frac{-6}{6003 \times 5997}
\]
Now substituting this back:
\[
E_2 - E_1 = 12400 \times \frac{-6}{6003 \times 5997}
\]
\[
= -\frac{74400}{6003 \times 5997}
\]
### Step 5: Approximate the denominator
Calculating \( 6003 \times 5997 \):
\[
6003 \times 5997 \approx 6000^2 = 36000000
\]
Thus,
\[
E_2 - E_1 \approx -\frac{74400}{36000000} \approx -0.00206667 \, \text{eV}
\]
### Final Result
The approximate energy difference between the two levels corresponding to \( \lambda_1 \) and \( \lambda_2 \) is:
\[
\approx 2 \times 10^{-3} \, \text{eV}
\]
