What will be the value of the longest wavelength line in Balmer's spectrum for H atom?

Options

A. 546 nm
B. 656 nm
C. 566 nm
D. 556nm

Answer

To find the value of the longest wavelength line in Balmer's spectrum for the hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Balmer Series**:
In the Balmer series, transitions occur from higher energy levels (n2) to the second energy level (n1 = 2). The possible values for n2 are 3, 4, 5, etc.

2. **Determine the Longest Wavelength**:
The longest wavelength corresponds to the smallest energy transition, which occurs when n2 is at its minimum value. Therefore, we set n2 = 3.

3. **Use the Rydberg Formula**:
The Rydberg formula for the wavelength (λ) of the emitted light is given by:
\[ \frac{1}{\lambda} = R_H \cdot z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant, \( z \) is the atomic number (for hydrogen, \( z = 1 \)), \( n_1 = 2 \), and \( n_2 = 3 \).

4. **Substitute the Values**:
Plugging in the values into the formula:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \, \text{m}^{-1} \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) \]

5. **Calculate the Fraction**:
Calculate \( \frac{1}{4} - \frac{1}{9} \):
\[ \frac{1}{4} = 0.25, \quad \frac{1}{9} \approx 0.1111 \]
\[ \frac{1}{4} - \frac{1}{9} = 0.25 - 0.1111 = 0.1389 \]

6. **Calculate \( \frac{1}{\lambda} \)**:
Now substitute back:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot 0.1389 \approx 1.527 \times 10^6 \, \text{m}^{-1} \]

7. **Calculate Wavelength \( \lambda \)**:
To find \( \lambda \), take the reciprocal:
\[ \lambda = \frac{1}{1.527 \times 10^6} \approx 6.54 \times 10^{-7} \, \text{m} \]
Converting to nanometers:
\[ \lambda \approx 654 \, \text{nm} \]

8. **Conclusion**:
The longest wavelength line in the Balmer spectrum for the hydrogen atom is approximately **656 nm**.