Assuming Rydberg’s constant (`R_H`) to be 109670 `cm^(–1)`, the longest wavelength line in the Lyman series of the hydrogen spectrum (in Å) is:

Explanation

To find the longest wavelength line in the Lyman series of the hydrogen spectrum, we can use the Rydberg formula for hydrogen:

\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_i^2} - \frac{1}{n_f^2} \right) \]

Where:
- \(\lambda\) is the wavelength,
- \(R_H\) is the Rydberg constant (given as \(109670 \, \text{cm}^{-1}\)),
- \(Z\) is the atomic number (for hydrogen, \(Z = 1\)),
- \(n_i\) is the initial energy level,
- \(n_f\) is the final energy level.

### Step 1: Identify the final state for the Lyman series
In the Lyman series, electrons transition to the \(n = 1\) level. Therefore, \(n_f = 1\).

### Step 2: Determine the initial state for the longest wavelength
To find the longest wavelength, we need to consider the transition from the highest possible initial state. The highest state that can transition to \(n = 1\) is \(n = 2\) (as \(n\) cannot be 1 for the initial state). Thus, \(n_i = 2\).

### Step 3: Substitute values into the Rydberg formula
Substituting \(Z = 1\), \(n_i = 2\), and \(n_f = 1\) into the Rydberg formula:

\[ \frac{1}{\lambda} = R_H \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{1^2} \right) \]

This simplifies to:

\[ \frac{1}{\lambda} = 109670 \cdot \left( \frac{1}{4} - 1 \right) \]

Calculating the expression inside the parentheses:

\[ \frac{1}{4} - 1 = \frac{1 - 4}{4} = -\frac{3}{4} \]

So we have:

\[ \frac{1}{\lambda} = 109670 \cdot \left(-\frac{3}{4}\right) \]

### Step 4: Calculate \(\lambda\)
Now, we can calculate \(\lambda\):

\[ \frac{1}{\lambda} = -\frac{3 \cdot 109670}{4} \]

Calculating this gives:

\[ \frac{1}{\lambda} = -\frac{329010}{4} = -82252.5 \, \text{cm}^{-1} \]

Since we are interested in the positive wavelength, we take the reciprocal:

\[ \lambda = \frac{4}{3 \cdot 109670} \]

Calculating this gives:

\[ \lambda \approx 1215.77 \times 10^{-8} \, \text{cm} \]

### Step 5: Convert to Angstroms
To convert centimeters to Angstroms, we use the conversion \(1 \, \text{cm} = 10^{10} \, \text{Å}\):

\[ \lambda = 1215.77 \times 10^{-8} \, \text{cm} \times 10^{10} \, \text{Å/cm} = 121.577 \, \text{Å} \]

### Final Answer
Thus, the longest wavelength line in the Lyman series of the hydrogen spectrum is approximately:

\[ \lambda \approx 121.577 \, \text{Å} \]