To find the value of the longest wavelength line in the Balmer spectrum for the hydrogen atom, we can follow these steps:
### Step 1: Understand the Balmer Series
The Balmer series corresponds to the transitions of electrons in a hydrogen atom from higher energy levels (n2) to the second energy level (n1 = 2). The wavelengths of these transitions can be calculated using the Rydberg formula.
### Step 2: Rydberg Formula
The Rydberg formula is given by:
\[
\frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\]
where:
- \( \lambda \) is the wavelength,
- \( R_H \) is the Rydberg constant (\( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level (for Balmer series, \( n_1 = 2 \)),
- \( n_2 \) is the higher energy level (can be 3, 4, 5, ...).
### Step 3: Longest Wavelength Condition
To find the longest wavelength, we need to minimize \( \frac{1}{\lambda} \). This occurs when \( n_2 \) is at its lowest value, which is \( n_2 = 3 \) (the first transition in the Balmer series).
### Step 4: Substitute Values into the Rydberg Formula
Substituting \( n_1 = 2 \) and \( n_2 = 3 \) into the Rydberg formula:
\[
\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right)
\]
Calculating the fractions:
\[
\frac{1}{2^2} = \frac{1}{4}, \quad \frac{1}{3^2} = \frac{1}{9}
\]
Now, find a common denominator (36):
\[
\frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36}
\]
Thus,
\[
\frac{1}{2^2} - \frac{1}{3^2} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36}
\]
### Step 5: Calculate \( \frac{1}{\lambda} \)
Now substitute back into the Rydberg equation:
\[
\frac{1}{\lambda} = 1.097 \times 10^7 \cdot \frac{5}{36}
\]
Calculating this gives:
\[
\frac{1}{\lambda} = 1.097 \times 10^7 \cdot 0.138888 \approx 1.528 \times 10^6 \, \text{m}^{-1}
\]
### Step 6: Calculate \( \lambda \)
Now, take the reciprocal to find \( \lambda \):
\[
\lambda = \frac{1}{1.528 \times 10^6} \approx 6.54 \times 10^{-7} \, \text{m}
\]
Convert meters to nanometers (1 m = \( 10^9 \) nm):
\[
\lambda \approx 654.3 \, \text{nm}
\]
### Step 7: Final Answer
The longest wavelength line in the Balmer spectrum for the hydrogen atom is approximately:
\[
\lambda \approx 656 \, \text{nm}
\]
