If the series limit wavelength of the Lyman series for hydrogen atom is `912 Å`, then the series limit wavelength for the Balmer series for the hydrogen atom is

Options

A. `912 Å//2`
B. `912 Å`
C. `912 xx 2 Å`
D. `912 xx 4 Å`

Explanation

To solve the problem of finding the series limit wavelength for the Balmer series given that the series limit wavelength for the Lyman series is 912 Å, we can follow these steps:

### Step 1: Understand the Series Limit
The series limit wavelength for a series in the hydrogen atom corresponds to the transition of an electron from a higher energy level to the lowest energy level. For the Lyman series, this transition is from \( n = \infty \) to \( n = 1 \).

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength \( \lambda \) of the emitted light is given by:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 3: Apply the Formula for the Lyman Series
For the Lyman series:
- \( n_1 = 1 \)
- \( n_2 = \infty \)

Substituting these values into the Rydberg formula gives:
\[ \frac{1}{\lambda_1} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \cdot 1 \cdot (1 - 0) = R \]
Thus, the series limit wavelength for the Lyman series is:
\[ \lambda_1 = \frac{1}{R} \]

### Step 4: Apply the Formula for the Balmer Series
For the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = \infty \)

Substituting these values into the Rydberg formula gives:
\[ \frac{1}{\lambda_2} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \cdot ( \frac{1}{4} - 0) = \frac{R}{4} \]
Thus, the series limit wavelength for the Balmer series is:
\[ \lambda_2 = \frac{4}{R} \]

### Step 5: Relate the Two Wavelengths
From the above calculations, we have:
\[ \lambda_1 = \frac{1}{R} \quad \text{and} \quad \lambda_2 = \frac{4}{R} \]
This implies:
\[ \lambda_2 = 4 \cdot \lambda_1 \]

### Step 6: Substitute the Given Value
Given that \( \lambda_1 = 912 \, \text{Å} \):
\[ \lambda_2 = 4 \cdot 912 \, \text{Å} = 3648 \, \text{Å} \]

### Final Answer
The series limit wavelength for the Balmer series for the hydrogen atom is \( 3648 \, \text{Å} \).

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