For the first member of Balmer series of hydrogen spectrum, the wavelength is `lamda`. What is the wavelength of the second member?
a.`(27)/(20) lambda` b.`(20)/(27)lambda` c.`(27)/(10) lambda` d.None of these

Options

A. `(27)/(20) lambda`
B. `(20)/(27)lambda`
C. `(27)/(10) lambda`
D. None of these

Explanation

To find the wavelength of the second member of the Balmer series of the hydrogen spectrum, we will follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to transitions where the electron falls to the second energy level (n1 = 2) from higher energy levels (n2 = 3, 4, 5,...). The first member of the Balmer series corresponds to the transition from n2 = 3 to n1 = 2.

### Step 2: Use the Formula for Wavelength
The formula for the wavelength of the emitted light during a transition in a hydrogen atom is given by:

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

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 upper energy levels, respectively.

### Step 3: Calculate for the First Member
For the first member of the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Substituting these values into the formula:

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

Calculating the right-hand side:

\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \cdot \frac{5}{36} \]

Thus,

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

### Step 4: Calculate for the Second Member
For the second member of the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = 4 \)

Substituting these values into the formula:

\[ \frac{1}{\lambda'} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

Calculating the right-hand side:

\[ \frac{1}{\lambda'} = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4 - 1}{16} \right) = R \cdot \frac{3}{16} \]

Thus,

\[ \lambda' = \frac{16}{3R} \]

### Step 5: Relate the Two Wavelengths
Now we have:
- \( \lambda = \frac{36}{5R} \)
- \( \lambda' = \frac{16}{3R} \)

To find the ratio \( \frac{\lambda}{\lambda'} \):

\[ \frac{\lambda}{\lambda'} = \frac{\frac{36}{5R}}{\frac{16}{3R}} = \frac{36 \cdot 3}{5 \cdot 16} = \frac{108}{80} = \frac{27}{20} \]

Thus, we can express \( \lambda' \) in terms of \( \lambda \):

\[ \lambda' = \lambda \cdot \frac{20}{27} \]

### Final Answer
The wavelength of the second member of the Balmer series is:

\[ \lambda' = \frac{20}{27} \lambda \]