Taking Rydberg's constant `R_(H) = 1.097 xx 10^(7)m` first and second wavelength of Balmer series in hydrogen spectrum is

Options

A. `2000 Å, 3000 Å`
B. `1575 Å, 2960 Å`
C. `6529 Å, 4280 Å`
D. `6552 Å, 4863 Å`

Explanation

To find the first and second wavelengths of the Balmer series in the hydrogen spectrum using Rydberg's formula, we will follow these steps:

### Step 1: Understand Rydberg's Formula
Rydberg's formula for the wavelengths of spectral lines in hydrogen is given by:

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

Where:
- \( \lambda \) is the wavelength,
- \( R_H \) is the Rydberg constant (\( R_H = 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the principal quantum number of the lower energy level,
- \( n_2 \) is the principal quantum number of the higher energy level.

### Step 2: Calculate the First Wavelength (n1 = 2, n2 = 3)
For the first wavelength in the Balmer series, we have:
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Substituting these values into Rydberg's formula:

\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

Calculating the right side:

\[ \frac{1}{\lambda_1} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) \]

Finding a common denominator (36):

\[ \frac{1}{\lambda_1} = 1.097 \times 10^7 \left( \frac{9 - 4}{36} \right) = 1.097 \times 10^7 \left( \frac{5}{36} \right) \]

Calculating \( \frac{5}{36} \):

\[ \frac{5}{36} \approx 0.1389 \]

Now substituting back:

\[ \frac{1}{\lambda_1} \approx 1.097 \times 10^7 \times 0.1389 \approx 1.527 \times 10^6 \]

Taking the reciprocal to find \( \lambda_1 \):

\[ \lambda_1 \approx \frac{1}{1.527 \times 10^6} \approx 6.54 \times 10^{-7} \, \text{m} = 654 \, \text{nm} \]

### Step 3: Calculate the Second Wavelength (n1 = 2, n2 = 4)
For the second wavelength in the Balmer series, we have:
- \( n_1 = 2 \)
- \( n_2 = 4 \)

Substituting these values into Rydberg's formula:

\[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

Calculating the right side:

\[ \frac{1}{\lambda_2} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right) \]

Finding a common denominator (16):

\[ \frac{1}{\lambda_2} = 1.097 \times 10^7 \left( \frac{4 - 1}{16} \right) = 1.097 \times 10^7 \left( \frac{3}{16} \right) \]

Calculating \( \frac{3}{16} \):

\[ \frac{3}{16} = 0.1875 \]

Now substituting back:

\[ \frac{1}{\lambda_2} \approx 1.097 \times 10^7 \times 0.1875 \approx 2.058 \times 10^6 \]

Taking the reciprocal to find \( \lambda_2 \):

\[ \lambda_2 \approx \frac{1}{2.058 \times 10^6} \approx 4.86 \times 10^{-7} \, \text{m} = 486 \, \text{nm} \]

### Final Results
- The first wavelength \( \lambda_1 \) is approximately **654 nm**.
- The second wavelength \( \lambda_2 \) is approximately **486 nm**.