The wavelength of spectral line in Lyman series for electron jumping back from 2nd orbit is

Options

A. `1162 Å`
B. `1216 Å`
C. `1362 Å`
D. `1176 Å`

Explanation

To find the wavelength of the spectral line in the Lyman series for an electron jumping from the second orbit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula**:
The formula to calculate the wavelength (λ) of spectral lines in the hydrogen atom is given by the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant.

2. **Define Variables**:
In the Lyman series, the electron transitions to the first orbit (n1 = 1). For the transition from the second orbit, we have:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

3. **Substitute Values**:
The Rydberg constant \( R_H \) is approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \) (or \( 1.097 \times 10^5 \, \text{cm}^{-1} \)).
Now substitute the values into the formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
\[ \frac{1}{\lambda} = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right) \]

4. **Calculate**:
Substitute \( R_H \):
\[ \frac{1}{\lambda} = 1.097 \times 10^5 \, \text{cm}^{-1} \times \frac{3}{4} \]
\[ \frac{1}{\lambda} = 1.097 \times 10^5 \times 0.75 = 8.2275 \times 10^4 \, \text{cm}^{-1} \]

5. **Find λ**:
Now, take the reciprocal to find λ:
\[ \lambda = \frac{1}{8.2275 \times 10^4} \approx 1.216 \times 10^{-5} \, \text{cm} \]

6. **Convert to Angstroms**:
Since \( 1 \, \text{cm} = 10^8 \, \text{Å} \):
\[ \lambda \approx 1.216 \times 10^{-5} \, \text{cm} \times 10^8 \, \text{Å/cm} = 1216 \, \text{Å} \]

### Final Answer:
The wavelength of the spectral line in the Lyman series for an electron jumping from the second orbit is approximately **1216 Å**.