Smallest Wavelength in Infrared Region of Hydrogen Spectrum

To find the smallest wavelength in the infrared region of the hydrogen spectrum, we use the Rydberg formula for the wavelength $\lambda$ of spectral lines:

$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

where $R$ is the Rydberg constant, and $n_1, n_2$ are principal quantum numbers with $n_2 > n_1$.

1. Ultraviolet (UV) Region:
   The UV region corresponds to the Lyman series, where $n_1 = 1$. The largest wavelength ($\lambda_{\text{max, UV}}$) occurs for the transition with the minimum energy, which corresponds to $n_2 = 2$:
   $\frac{1}{\lambda_{\text{max, UV}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3}{4}R$

Given that $\lambda_{\text{max, UV}} = 122\text{ nm}$:
$\frac{1}{122} = \frac{3}{4}R \implies R = \frac{4}{3 \times 122}\text{ nm}^{-1}$

1. Infrared (IR) Region:
   The infrared region consists of the Paschen series ($n_1 = 3$), Brackett series ($n_1 = 4$), Pfund series ($n_1 = 5$), etc.
   The smallest wavelength in the infrared spectrum corresponds to the transition with the highest energy in the series with the lowest $n_1$ in this region, which is the series limit ($n_2 = \infty$) of the Paschen series ($n_1 = 3$):
   $\frac{1}{\lambda_{\text{min, IR}}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = \frac{R}{9}$

Therefore:
$\lambda_{\text{min, IR}} = \frac{9}{R}$

Substitute the value of $R$:
$\lambda_{\text{min, IR}} = 9 \times \frac{3 \times 122}{4} = \frac{27 \times 122}{4} = \frac{3294}{4} = 823.5\text{ nm} \approx 823\text{ nm}$

Hence, the correct option is B.