The shortest wavelength present in the Lyman series of spectral lines is: (Given Rydberg constant R = 1.097 x 107 m^-1)

The correct option is:
$\begin{array}{c}9.1\times {10}^{-8}\text{m}\end{array}$

Step-by-Step Explanation:

1. Understanding the Lyman Series:
The Lyman series corresponds to spectral lines emitted when an electron in a hydrogen atom undergoes a transition from an outer orbit (where the principal quantum number n₂ = 2, 3, 4, ...) to the innermost orbit (where the principal quantum number n₁ = 1).

The wavelength $\lambda$ of the emitted spectral line is given by the Rydberg formula:
$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
Here, R is the Rydberg constant, which is given as:
$R=1.097\times {10}^7{\text{m}}^{-1}$

2. Finding the Condition for the Shortest Wavelength:
For the Lyman series, we substitute n₁ = 1:
$\frac{1}{\lambda }=R\left(1-\frac{1}{n_2^2}\right)$
The shortest wavelength (also called the series limit) corresponds to the maximum energy transition, which occurs when the electron transitions from an orbit at infinity (n₂ = ∞) to the ground state (n₁ = 1).

Substituting $n_2=\infty$ into the Rydberg formula:
$\frac{1}{{\lambda }_{\text{min}}}=R\left(1-\frac{1}{{\infty }^2}\right)$
Since $\frac{1}{{\infty }^2}=0$, we have:
$\frac{1}{{\lambda }_{\text{min}}}=R$
Therefore, the shortest wavelength is:
${\lambda }_{\text{min}}=\frac{1}{R}$

3. Calculating the Value:
Substituting the given value of R:
${\lambda }_{\text{min}}=\frac{1}{1.097\times {10}^7}\text{m}$
${\lambda }_{\text{min}}\approx 0.9116\times {10}^{-7}\text{m}$
${\lambda }_{\text{min}}\approx 9.1\times {10}^{-8}\text{m}$

Thus, the shortest wavelength present in the Lyman series is approximately 9.1 × 10^-8 m.