The shortest wavelength in hydrogen spectrum of Lyman series
when `R_(H)=109678 cm^(-1)` is :-

Options

A. `1002.7overset(@)A`
B. `1215.67overset(@)A`
C. `1127.30overset(@)A`
D. `911.7overset(@)A`

Answer

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

### Step-by-Step Solution:

1. **Identify the Formula**:
The formula for the wavelength in the hydrogen spectrum is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant for hydrogen, \( n_1 \) is the lower energy level, and \( n_2 \) is the higher energy level.

2. **Determine \( n_1 \) and \( n_2 \)**:
For the Lyman series, the transition starts from \( n_2 \) to \( n_1 \):
- The lowest energy level \( n_1 \) for the Lyman series is 1.
- The maximum energy level \( n_2 \) approaches infinity (as we want the shortest wavelength).

3. **Substitute Values**:
Substitute \( n_1 = 1 \) and \( n_2 = \infty \) into the formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies to:
\[ \frac{1}{\lambda} = R_H \left( 1 - 0 \right) = R_H \]

4. **Insert the Value of \( R_H \)**:
Given \( R_H = 109678 \, \text{cm}^{-1} \):
\[ \frac{1}{\lambda} = 109678 \]

5. **Calculate \( \lambda \)**:
To find \( \lambda \), take the reciprocal:
\[ \lambda = \frac{1}{109678} \, \text{cm} \]
Calculating this gives:
\[ \lambda \approx 9.1170 \times 10^{-6} \, \text{cm} \]

6. **Convert to Angstroms**:
Since 1 Angstrom = \( 10^{-8} \, \text{cm} \), we convert:
\[ \lambda = 9.1170 \times 10^{-6} \, \text{cm} \times \frac{10^{10} \, \text{Å}}{1 \, \text{cm}} = 911.7 \, \text{Å} \]

7. **Final Answer**:
The shortest wavelength in the hydrogen spectrum of the Lyman series is:
\[ \lambda \approx 911.7 \, \text{Å} \]