The first line in the Balmer series in the H atom will have the frequency

Options

A. `4.57 xx 10^(14)s^(-1)`
B. `3.29 xx 10^(15) s^(-1)`
C. `8.22 xx 10^(15) s^(-1)`
D. `8.05 xx 10^(13) s^(-1)`

Explanation

To find the frequency of the first line in the Balmer series of the hydrogen atom, we can follow these steps:

### Step 1: Identify the Quantum Levels
In the Balmer series, the transitions occur from higher energy levels (N2) to the second energy level (N1 = 2). For the first line of the Balmer series, the transition is from N2 = 3 to N1 = 2.

### Step 2: Use the Rydberg Formula
The Rydberg formula for hydrogen is given by:

\[ \frac{1}{\lambda} = R \left( \frac{1}{N_1^2} - \frac{1}{N_2^2} \right) \]

Where:
- \( R \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)
- \( N_1 = 2 \)
- \( N_2 = 3 \)

### Step 3: Substitute the Values
Substituting \( N_1 \) and \( N_2 \) into the formula:

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

Calculating the squares:

\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

### Step 4: Find a Common Denominator
The common denominator for 4 and 9 is 36:

\[ \frac{1}{\lambda} = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]

### Step 5: Substitute the Rydberg Constant
Now substituting \( R \):

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{5}{36} \right) \]

### Step 6: Calculate \( \frac{1}{\lambda} \)
Calculating \( \frac{1}{\lambda} \):

\[ \frac{1}{\lambda} = \frac{1.097 \times 10^7 \times 5}{36} \]

Calculating this gives:

\[ \frac{1}{\lambda} \approx 1.52 \times 10^6 \, \text{m}^{-1} \]

### Step 7: Convert Wavelength to Frequency
Using the relationship between frequency (\( \nu \)), wavelength (\( \lambda \)), and the speed of light (\( c \)):

\[ c = \nu \lambda \implies \nu = \frac{c}{\lambda} \]

Substituting \( \frac{1}{\lambda} \):

\[ \nu = c \cdot \frac{1}{\lambda} = c \cdot \frac{1.52 \times 10^6}{1} \]

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

\[ \nu = 3 \times 10^8 \times 1.52 \times 10^6 \]

### Step 8: Final Calculation
Calculating this gives:

\[ \nu \approx 4.57 \times 10^{14} \, \text{s}^{-1} \]

### Conclusion
The frequency of the first line in the Balmer series of the hydrogen atom is approximately \( 4.57 \times 10^{14} \, \text{s}^{-1} \).

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