Calculate wavelength of the radiation corresponding to the speciral line of the lowest frequency in lyman series in the spectrum of a hydrogen atom `(R_(H) = 109677 cm^(-1), c = 3 xx 10^(8) m s^(-1), Z = 1)`

Explanation

To calculate the wavelength of the radiation corresponding to the spectral line of the lowest frequency in the Lyman series of the hydrogen atom, we can follow these steps:

### Step 1: Identify the values for n1 and n2
In the Lyman series, the transitions occur from higher energy levels (n2) to the first energy level (n1 = 1). The lowest frequency transition corresponds to the transition from n2 = 2 to n1 = 1.

- **n1 = 1**
- **n2 = 2**

### Step 2: Use the Rydberg formula
The Rydberg formula for the wavelength (λ) of the emitted radiation is given by:

\[ \frac{1}{\lambda} = R_Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( R \) is the Rydberg constant for hydrogen, which is given as \( R_H = 109677 \, \text{cm}^{-1} \).
- \( Z \) is the atomic number of hydrogen, which is 1.

### Step 3: Substitute the values into the formula
Substituting the values into the Rydberg formula:

\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Substituting the known values:

\[ \frac{1}{\lambda} = 109677 \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

Calculating the fractions:

\[ \frac{1}{\lambda} = 109677 \left( 1 - \frac{1}{4} \right) = 109677 \left( \frac{3}{4} \right) \]

### Step 4: Calculate \( \frac{1}{\lambda} \)
Now, calculate \( \frac{1}{\lambda} \):

\[ \frac{1}{\lambda} = 109677 \cdot \frac{3}{4} = 82008.75 \, \text{cm}^{-1} \]

### Step 5: Calculate the wavelength \( \lambda \)
To find \( \lambda \), take the reciprocal of \( \frac{1}{\lambda} \):

\[ \lambda = \frac{1}{82008.75} \, \text{cm} \]

Calculating \( \lambda \):

\[ \lambda \approx 1.219 \times 10^{-5} \, \text{cm} \]

### Step 6: Convert to nanometers
To convert from centimeters to nanometers (1 cm = \( 10^7 \) nm):

\[ \lambda \approx 1.219 \times 10^{-5} \, \text{cm} \times 10^7 \, \text{nm/cm} = 121.9 \, \text{nm} \]

### Final Answer
The wavelength of the radiation corresponding to the spectral line of the lowest frequency in the Lyman series is approximately:

\[ \lambda \approx 121.9 \, \text{nm} \]