The wavelength of the first line of Balmer series is `6563 Å`. The Rydberg's constant for hydrogen is about

Options

A. `1.09 xx 10^(5) m^(-1)`
B. `1.09 xx 10^(6) m^(-1)`
C. `1.097 xx 10^(7) m^(-1)`
D. `1.09 xx 10^(8) m^(-1)`

Explanation

To find the Rydberg constant for hydrogen using the wavelength of the first line of the Balmer series, we can follow these steps:

### Step 1: Understand the given information
We are given the wavelength of the first line of the Balmer series, which is \( \lambda = 6563 \, \text{Å} \). We need to convert this wavelength into meters for our calculations.

### Step 2: Convert the wavelength to meters
1 Ångström (Å) is equal to \( 10^{-10} \) meters. Therefore:
\[ \lambda = 6563 \, \text{Å} = 6563 \times 10^{-10} \, \text{m} \]

### Step 3: Use the Rydberg formula
The Rydberg formula for the wavelength of spectral lines in hydrogen is given by:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the Balmer series, the transitions are from higher energy levels to \( n_1 = 2 \). The first line corresponds to \( n_2 = 3 \).

### Step 4: Substitute the values into the formula
Substituting \( n_1 = 2 \) and \( n_2 = 3 \):
\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating \( \frac{1}{2^2} \) and \( \frac{1}{3^2} \):
\[ \frac{1}{2^2} = \frac{1}{4} \quad \text{and} \quad \frac{1}{3^2} = \frac{1}{9} \]
Now, find a common denominator (which is 36):
\[ \frac{1}{4} = \frac{9}{36} \quad \text{and} \quad \frac{1}{9} = \frac{4}{36} \]
Thus:
\[ \frac{1}{2^2} - \frac{1}{3^2} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

### Step 5: Substitute back into the Rydberg formula
Now we can rewrite the formula:
\[ \frac{1}{\lambda} = R \left( \frac{5}{36} \right) \]
Rearranging gives:
\[ R = \frac{36}{5 \lambda} \]

### Step 6: Substitute the value of \( \lambda \)
Now substitute \( \lambda = 6563 \times 10^{-10} \, \text{m} \):
\[ R = \frac{36}{5 \times 6563 \times 10^{-10}} \]

### Step 7: Calculate the Rydberg constant
Calculating the value:
\[ R = \frac{36}{5 \times 6563 \times 10^{-10}} \approx \frac{36}{32815 \times 10^{-10}} \approx 1.095 \times 10^7 \, \text{m}^{-1} \]
Thus, rounding gives:
\[ R \approx 1.09 \times 10^7 \, \text{m}^{-1} \]

### Final Answer
The Rydberg constant for hydrogen is approximately \( R \approx 1.09 \times 10^7 \, \text{m}^{-1} \).

---