Step 1: Approach
The highest energy photon in a series is emitted when the electron falls from infinity. That photon has the smallest wavelength.
Step 2: Wavenumber
The Lyman series ends on $n = 1$, so the wavenumber is $\bar{\nu} = R_H/1^2 = 109677\ \text{cm}^{-1}$.
Step 3: Wavelength
Take the reciprocal: $\lambda = 1/\bar{\nu}$.
\[ \lambda = \frac{1}{109677}\ \text{cm} \approx 9.117 \times 10^{-6}\ \text{cm} \]
Step 4: Sanity check
This is about 91.2 nm, the well known Lyman limit in the ultraviolet. A transition such as $n = 2 \to 1$ gives 121.6 nm, which is longer, so it is not the shortest.
Final Answer:
The shortest Lyman wavelength is about 91.2 nm. This is option (A).
\[ \boxed{\text{(A) }9.117 \times 10^{-6} \ \text{cm}} \]