Wavelength of the first line of balmer seris is 600 nm. The wavelength of second line of the balmer series will be

Options

A. 444 nm
B. 800 nm
C. 388 nm
D. 632 nm

Answer

To find the wavelength of the second line of the Balmer series, we can follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to the transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 3) to the second energy level (n = 2). The first line of the Balmer series corresponds to the transition from n = 3 to n = 2, and the second line corresponds to the transition from n = 4 to n = 2.

### Step 2: Use the Rydberg Formula
The wavelength of light emitted during these transitions can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy levels, respectively.

### Step 3: Calculate for the First Line of the Balmer Series
For the first line (n = 3 to n = 2):
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_1} = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]

Given that \( \lambda_1 = 600 \, \text{nm} \), we can express this as:
\[ \frac{1}{600} = R \left( \frac{5}{36} \right) \]

### Step 4: Calculate for the Second Line of the Balmer Series
For the second line (n = 4 to n = 2):
- \( n_1 = 2 \)
- \( n_2 = 4 \)

Substituting these values into the Rydberg formula:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) \]
Finding a common denominator (16):
\[ \frac{1}{\lambda_2} = R \left( \frac{4 - 1}{16} \right) = R \left( \frac{3}{16} \right) \]

### Step 5: Relate the Two Wavelengths
Now we can relate the two wavelengths:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\frac{R \cdot \frac{3}{16}}{R \cdot \frac{5}{36}}} = \frac{3}{16} \cdot \frac{36}{5} = \frac{3 \cdot 36}{16 \cdot 5} = \frac{108}{80} = \frac{27}{20} \]
Thus,
\[ \lambda_2 = \lambda_1 \cdot \frac{27}{20} \]

### Step 6: Calculate the Wavelength of the Second Line
Substituting \( \lambda_1 = 600 \, \text{nm} \):
\[ \lambda_2 = 600 \cdot \frac{27}{20} = 600 \cdot 1.35 = 810 \, \text{nm} \]

### Final Result
The wavelength of the second line of the Balmer series is approximately **810 nm**.

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