The shortest wavelength in the Lyman series of hydrogen spectrum is `912 Å` correcponding to a photon energy of `13.6eV`. The shortest wavelength in the Balmer series is about

Options

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

Explanation

To find the shortest wavelength in the Balmer series of the hydrogen spectrum, we can follow these steps:

### Step 1: Understand the relationship between wavelength and energy
We know that the energy of a photon is given by the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength.

### Step 2: Identify the shortest wavelength in the Lyman series
The shortest wavelength in the Lyman series is given as \( 912 \, \text{Å} \) (or \( 912 \times 10^{-10} \, \text{m} \)). This corresponds to the transition from \( n = \infty \) to \( n = 1 \).

### Step 3: Use the Rydberg formula for the Balmer series
The Rydberg formula for the wavelengths of spectral lines in hydrogen is:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant (\( 1.097 \times 10^7 \, \text{m}^{-1} \)), \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

For the Balmer series, the transitions start from \( n_2 \) (where \( n_2 \geq 2 \)) to \( n_1 = 2 \).

### Step 4: Calculate the shortest wavelength in the Balmer series
The shortest wavelength in the Balmer series corresponds to the transition from \( n_2 = \infty \) to \( n_1 = 2 \):
\[ \frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]
\[ \frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{4} - 0 \right) \]
\[ \lambda_{\text{min}} = \frac{4}{R} \]

### Step 5: Relate the Balmer series to the Lyman series
From the Lyman series, we know:
\[ \lambda_{\text{min, Lyman}} = \frac{1}{R} \]
Thus, for the Balmer series:
\[ \lambda_{\text{min, Balmer}} = 4 \times \lambda_{\text{min, Lyman}} \]
\[ \lambda_{\text{min, Balmer}} = 4 \times 912 \, \text{Å} \]

### Step 6: Calculate the final answer
\[ \lambda_{\text{min, Balmer}} = 3648 \, \text{Å} \]

### Final Answer
The shortest wavelength in the Balmer series is approximately \( 3648 \, \text{Å} \).

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