Calculate (a) the wavelength and (b) the frequencey of the `Hbeta`
line of the Balmer series for hydrogen.

Explanation

To solve the problem of calculating the wavelength and frequency of the H-beta line of the Balmer series for hydrogen, we will follow these steps:

### Step 1: Identify the Transition
The H-beta line corresponds to the transition from n=4 to n=2 in the hydrogen atom.

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

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

Where:
- \( R \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \))
- \( n_1 \) is the lower energy level (2 for H-beta)
- \( n_2 \) is the higher energy level (4 for H-beta)

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

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

Calculating the fractions:

\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{4^2} = \frac{1}{16} = 0.0625 \]

Now substituting these values:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( 0.25 - 0.0625 \right) \]
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.1875 \]
\[ \frac{1}{\lambda} = 0.2053125 \times 10^7 \, \text{m}^{-1} \]

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

\[ \lambda = \frac{1}{0.2053125 \times 10^7} \approx 4.87 \times 10^{-7} \, \text{m} \]

### Step 5: Calculate Frequency (f)
The frequency (f) can be calculated using the speed of light (c) and the wavelength (λ):

\[ f = \frac{c}{\lambda} \]

Where \( c \approx 3 \times 10^8 \, \text{m/s} \).

Substituting the values:

\[ f = \frac{3 \times 10^8}{4.87 \times 10^{-7}} \approx 6.15 \times 10^{14} \, \text{Hz} \]

### Final Answers
(a) The wavelength of the H-beta line is approximately \( 4.87 \times 10^{-7} \, \text{m} \) (or 487 nm).

(b) The frequency of the H-beta line is approximately \( 6.15 \times 10^{14} \, \text{Hz} \).

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