The first emission line in the H-atom spectrum in the Balmer series appears at:

Options

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

Answer

To determine the first emission line in the hydrogen atom spectrum in the Balmer series, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Balmer Series**:
- The Balmer series corresponds to transitions where the electron falls to the second energy level (n1 = 2) from higher energy levels (n2 = 3, 4, 5, ...).

2. **Determine the First Emission Line**:
- The first emission line in the Balmer series corresponds to the transition from n2 = 3 to n1 = 2. This is the lowest energy transition that results in an emission line in the Balmer series.

3. **Use the Rydberg Formula**:
- The Rydberg formula for the wave number (ν̅) of the emitted light is given by:
\[ \nu̅ = R_H \left( \frac{Z^2}{n_1^2} - \frac{Z^2}{n_2^2} \right) \]
- For hydrogen (Z = 1), the formula simplifies to:
\[ \nu̅ = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

4. **Substitute the Values**:
- Substitute n1 = 2 and n2 = 3 into the formula:
\[ \nu̅ = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
- This becomes:
\[ \nu̅ = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]

5. **Calculate the Difference**:
- Find a common denominator (36):
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
- Therefore:
\[ \nu̅ = R_H \left( \frac{9}{36} - \frac{4}{36} \right) = R_H \left( \frac{5}{36} \right) \]

6. **Final Result**:
- The first emission line in the Balmer series appears at:
\[ \nu̅ = \frac{5 R_H}{36} \]
- The units for wave number (ν̅) are in cm⁻¹.

### Conclusion:
The first emission line in the hydrogen atom spectrum in the Balmer series appears at \(\frac{5 R_H}{36}\) cm⁻¹.