If longest wavelength of Balmer series of H atom is `lambda` then shortest wavelength of Lyman series will be

Options

A. `5/36 lambda`
B. `5/9 lambda`
C. `36/5 lambda`
D. `5/9 lambda`

Answer

To solve the problem, we need to find the relationship between the longest wavelength of the Balmer series and the shortest wavelength of the Lyman series for a hydrogen atom. Let's break this down step by step.

### Step 1: Understand the Series
- **Lyman Series**: Transitions from higher energy levels (n ≥ 2) to n = 1.
- **Balmer Series**: Transitions from higher energy levels (n ≥ 3) to n = 2.

### Step 2: Identify the Longest Wavelength of the Balmer Series
The longest wavelength in the Balmer series occurs when the transition is from n = 3 to n = 2.

Using the formula for wavelength:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = 3 \)
- \( Z = 1 \) (for hydrogen)

Substituting these values in:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Calculating the right side:
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Thus,
\[ \frac{1}{\lambda} = R \cdot \frac{5}{36} \]
Rearranging gives:
\[ \lambda = \frac{36}{5R} \]
Let’s denote this wavelength as \( \lambda = \lambda_B \).

### Step 3: Identify the Shortest Wavelength of the Lyman Series
The shortest wavelength in the Lyman series occurs when the transition is from n = ∞ to n = 1.

Using the same formula:
For the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = \infty \)

Substituting these values in:
\[ \frac{1}{\lambda_0} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \left( 1 - 0 \right) = R \]
Thus,
\[ \lambda_0 = \frac{1}{R} \]

### Step 4: Relate the Two Wavelengths
From the Balmer series, we found that:
\[ R = \frac{36}{5\lambda_B} \]
Substituting this into the expression for the shortest wavelength of the Lyman series:
\[ \lambda_0 = \frac{1}{R} = \frac{1}{\frac{36}{5\lambda_B}} = \frac{5\lambda_B}{36} \]

### Final Answer
Thus, the shortest wavelength of the Lyman series in terms of the longest wavelength of the Balmer series is:
\[ \lambda_0 = \frac{5}{36} \lambda_B \]