Calculate the wavelength of the first line in the Balmer series of hydrogen spectrum.

Explanation

To calculate the wavelength of the first line in the Balmer series of the hydrogen spectrum, we can follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to electron transitions in hydrogen where the final energy level (n_l) is 2. The first line in this series corresponds to the transition from n_h = 3 to n_l = 2.

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

\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_l^2} - \frac{1}{n_h^2} \right) \]

Where:
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \) or \( 109678 \, \text{cm}^{-1} \))
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \))
- \( n_l \) is the lower energy level (2 for the Balmer series)
- \( n_h \) is the higher energy level (3 for the first line in the Balmer series)

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

\[ \frac{1}{\lambda} = 109678 \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculating the fractions:

\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{3^2} = \frac{1}{9} \approx 0.1111 \]

Now, substituting these values back into the equation:

\[ \frac{1}{\lambda} = 109678 \left( 0.25 - 0.1111 \right) \]
\[ \frac{1}{\lambda} = 109678 \left( 0.1389 \right) \]
\[ \frac{1}{\lambda} \approx 15287.78 \, \text{cm}^{-1} \]

### Step 4: Calculate Wavelength (λ)
Now, take the reciprocal to find λ:

\[ \lambda = \frac{1}{15287.78} \approx 0.0000654 \, \text{cm} \]

### Step 5: Convert to Angstroms
To convert centimeters to angstroms (1 cm = \( 10^{10} \) angstroms):

\[ \lambda \approx 0.0000654 \, \text{cm} \times 10^{10} \, \text{angstroms/cm} = 654 \, \text{angstroms} \]

### Final Answer
The wavelength of the first line in the Balmer series of the hydrogen spectrum is approximately **6564 angstroms**.

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