The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at`

Options

A. `9R//400cm^(-1)`
B. `7R//144cm^(-1)`
C. `3R//4cm^(-1)`
D. `5R//36cm^(-1)`

Answer

To solve the question about the first emission line in the atomic spectrum of hydrogen in the Balmer series, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Balmer Series**:
- The Balmer series corresponds to the transitions of electrons from higher energy levels (n ≥ 3) to the second energy level (n = 2) in a hydrogen atom. The first line of the Balmer series occurs when an electron transitions from n = 3 to n = 2.

2. **Identify the Transition**:
- For the first line of the Balmer series:
- The initial energy level (n2) is 3.
- The final energy level (n1) is 2.

3. **Using the Rydberg Formula**:
- The Rydberg formula for calculating the wavelength (or wave number) of spectral lines is given by:
\[ \bar{\nu} = R_H \left( Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \right) \]
- Where:
- \(\bar{\nu}\) = wave number (in cm⁻¹)
- \(R_H\) = Rydberg constant (approximately 1.097 × 10⁷ m⁻¹)
- \(Z\) = atomic number (for hydrogen, \(Z = 1\))
- \(n_1\) = lower energy level (2)
- \(n_2\) = higher energy level (3)

4. **Substituting Values**:
- Substitute \(Z = 1\), \(n_1 = 2\), and \(n_2 = 3\) into the Rydberg formula:
\[ \bar{\nu} = R_H \left( 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \right) \]
- Calculate \(\frac{1}{2^2} = \frac{1}{4}\) and \(\frac{1}{3^2} = \frac{1}{9}\):
\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{4} - \frac{1}{9} \]
- To subtract these fractions, find a common denominator (36):
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
\[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

5. **Final Calculation**:
- Substitute back into the formula:
\[ \bar{\nu} = R_H \left( \frac{5}{36} \right) \]
- Using \(R_H \approx 1.097 \times 10^7 \, \text{m}^{-1}\):
\[ \bar{\nu} = 1.097 \times 10^7 \times \frac{5}{36} \approx 1.52 \times 10^6 \, \text{m}^{-1} \]
- Convert to cm⁻¹ (1 m = 100 cm):
\[ \bar{\nu} \approx 1.52 \times 10^4 \, \text{cm}^{-1} \]

6. **Conclusion**:
- The first emission line in the atomic spectrum of hydrogen in the Balmer series appears at approximately \(1.52 \times 10^4 \, \text{cm}^{-1}\).