Animated Solution for Physics - Atoms and Nuclei: The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is

In this problem, we are taking a journey through the spectral series of the hydrogen atom. We are given information about the ultraviolet region and asked to find a specific wavelength in the infrared region. Let's break down the physics and the math step-by-step.

The hydrogen emission spectrum is divided into several series based on the final energy level ($n_f$) the electron falls into. - The Ultraviolet (UV) region corresponds to the Lyman series, where electrons fall to the ground state, $n_f=1$. - The Infrared (IR) region begins with the Paschen series, where electrons fall to $n_f=3$.

The wavelength of any spectral line is governed by the Rydberg formula:

Here is a crucial conceptual bridge: Energy and wavelength are inversely proportional ($E=\frac{hc}{\lambda }$). - The largest wavelength (${\lambda }_{\text{max}}$) corresponds to the minimum energy jump, which happens when an electron falls from the immediately higher level ($n_i=n_f+1$). - The smallest wavelength (${\lambda }_{\text{min}}$) corresponds to the maximum energy jump, which happens when an electron falls from infinity ($n_i=\infty$).

Let's set up the equation for the largest wavelength in the UV region (Lyman series). Here, $n_f=1$ and $n_i=2$. We are given that ${\lambda }_{\text{UV, max}}=122\text{ nm}$.

Now, let's set up the equation for the smallest wavelength in the IR region (Paschen series). Here, $n_f=3$ and $n_i=\infty$.

To find $\lambda$, we can simply divide Equation 1 by Equation 2. This elegantly cancels out the Rydberg constant $R$, saving us from messy calculations.

Now, isolate $\lambda$:

The question asks for the answer to the nearest integer. Rounding $823.5\text{ nm}$ gives us $823\text{ nm}$, which corresponds to option (b).