Hydrogen Atom and Hydrogen Molecule
The observed wavelengths in the line spectrum of hydrogen atom were first expressed in terms of a series by johann Jakob Balmer, a Swiss teacher. Balmer's empirical empirical formula is
`1/lambda=R_(H)(1/2^(2)-1/n^(2))," " N=3, 4, 5`
`R_(H)=(Me e^(4))/(8 epsilon_(0)^(2)h^(3) c)=109. 678 cm^(-1)`
Here,
`R_(H)` is the Rydberg Constant, `m_(e)` is the mass of electron. Niels Bohr derived this expression theoretically in 1913. The formula is easily generalized to any one electron atom//ion. the wavelength of first line of balmer region.

Answer

To solve the problem regarding the wavelength of the first line of the Balmer region for the hydrogen atom, we can follow these steps:

### Step 1: Understand the Formula
The formula provided is:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \]
where \( R_H \) is the Rydberg constant, \( \lambda \) is the wavelength, and \( n \) is the principal quantum number of the electron's orbit.

### Step 2: Identify the Values
For the first line of the Balmer series, we need to set \( n = 3 \) (since the first line corresponds to the transition from \( n = 3 \) to \( n = 2 \)). We also have:
\[ R_H = 109678 \, \text{cm}^{-1} \]

### Step 3: Substitute the Values into the Formula
Substituting \( n = 3 \) into the formula gives:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating the individual terms:
\[ \frac{1}{2^2} = \frac{1}{4} \quad \text{and} \quad \frac{1}{3^2} = \frac{1}{9} \]
Thus,
\[ \frac{1}{\lambda} = 109678 \left( \frac{1}{4} - \frac{1}{9} \right) \]

### Step 4: Calculate the Difference
Now, we need to calculate \( \frac{1}{4} - \frac{1}{9} \):
\[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

### Step 5: Substitute Back into the Formula
Now substituting this back into our equation:
\[ \frac{1}{\lambda} = 109678 \times \frac{5}{36} \]

### Step 6: Calculate \( \frac{1}{\lambda} \)
Calculating \( 109678 \times \frac{5}{36} \):
\[ \frac{1}{\lambda} = \frac{548390}{36} \approx 15232.5 \, \text{cm}^{-1} \]

### Step 7: Find \( \lambda \)
To find \( \lambda \), take the reciprocal:
\[ \lambda = \frac{1}{15232.5} \approx 0.0000655 \, \text{cm} \]
Converting this to angstroms (1 cm = \( 10^8 \) angstroms):
\[ \lambda \approx 6569 \, \text{Å} \]

### Final Answer
The wavelength of the first line of the Balmer region is approximately:
\[ \lambda \approx 6569 \, \text{Å} \]