What is the maximum wavelength of line of Balmer series of hydrogen spectrum `(R=1.09xx10^(7)m^(-1))`-

Options

A. 912 A
B. 3645 A
C. 6561 A
D. 8201 A

Answer

To find the maximum wavelength of the Balmer series of the hydrogen spectrum, we can use the Rydberg formula:

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

where:
- \(\lambda\) is the wavelength,
- \(R\) is the Rydberg constant (\(1.09 \times 10^7 \, \text{m}^{-1}\)),
- \(n_1\) is the lower energy level,
- \(n_2\) is the higher energy level.

### Step 1: Identify the values of \(n_1\) and \(n_2\)
For the Balmer series, the lower energy level \(n_1\) is 2. The maximum wavelength corresponds to the transition from \(n_2 = 3\) to \(n_1 = 2\).

### Step 2: Substitute the values into the Rydberg formula
Substituting \(n_1 = 2\) and \(n_2 = 3\) into the Rydberg formula:

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

### Step 3: Calculate the fractions
Calculate \( \frac{1}{2^2} \) and \( \frac{1}{3^2} \):

\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{3^2} = \frac{1}{9} \approx 0.1111 \]

### Step 4: Find the difference
Now, find the difference:

\[ \frac{1}{4} - \frac{1}{9} = 0.25 - 0.1111 = \frac{9 - 4}{36} = \frac{5}{36} \]

### Step 5: Substitute back into the formula
Now substitute back into the Rydberg formula:

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

### Step 6: Solve for \(\lambda\)
Rearranging gives:

\[ \lambda = \frac{36}{5R} \]

Substituting \(R = 1.09 \times 10^7 \, \text{m}^{-1}\):

\[ \lambda = \frac{36}{5 \times 1.09 \times 10^7} \]

### Step 7: Calculate \(\lambda\)
Calculating this gives:

\[ \lambda = \frac{36}{5.45 \times 10^7} \approx 6.563 \times 10^{-7} \, \text{m} \]

### Step 8: Convert to nanometers
To convert meters to nanometers:

\[ \lambda \approx 6.563 \times 10^{-7} \, \text{m} = 6563 \, \text{nm} \]

### Conclusion
Thus, the maximum wavelength of the line of the Balmer series of the hydrogen spectrum is approximately \(6563 \, \text{nm}\).

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