Problem 44 - Chemistry: The Central Science in SI Units, Global Edition

Question

The hydrogen atom can absorb light of wavelength \(1094 \mathrm{nm}\). (a) In what region of the electromagnetic spectrum is this absorption found? (b) Determine the initial and final values of \(n\) associated with this absorption.

Step-by-Step Solution

Answer

(a) The absorption of light with a wavelength of \(1094 \mathrm{nm}\) occurs in the Infrared region of the electromagnetic spectrum. (b) The initial value of the principal quantum number (n) associated with this absorption is \(n_1 = 1\), and the final value of n is \(n_2 = 5\).

Step 1: (a) Identifying the region of the electromagnetic spectrum)

To determine the region of the electromagnetic spectrum where light with a wavelength of \(1094 \mathrm{nm}\) belongs, we can refer to the classification of the electromagnetic spectrum: - Radio waves: \(10^3\) to \(10^5 \mathrm{nm}\) - Microwaves: \(10^2\) to \(10^3 \mathrm{nm}\) - Infrared (IR): \(7 \times 10^2\) to \(10^5 \mathrm{nm}\) - Visible: \(400\) to \(700 \mathrm{nm}\) - Ultraviolet (UV): \(10\) to \(400 \mathrm{n}\) - X-rays: \(0.01\) to \(10 \mathrm{nm}\) - Gamma rays: < \(0.01 \mathrm{nm}\) Since \(1094 \mathrm{nm}\) falls in the range of the Infrared region (\(700\) to \(10^5 \mathrm{nm}\)), the absorption occurs in the Infrared region of the electromagnetic spectrum.

Step 2: (b) Determining initial and final values of n)

To find the values of the initial and final principle quantum numbers (n) associated with this absorption, we can use the Balmer-Rydberg equation: \(\frac{1}{\lambda} = R_H \left( \frac{1}{n_{1}^2} - \frac{1}{n_{2}^2} \right)\) Where \(\lambda\) is the wavelength of absorbed light, \(R_H\) is the Rydberg constant for hydrogen (\(1.097 \times 10^7 \mathrm{m^{-1}}\)), and \(n_1\) and \(n_2\) are the initial and final principal quantum numbers, respectively. Given the wavelength \(\lambda = 1094 \mathrm{nm} = 1.094 \times 10^{-6} \mathrm{m}\), we can plug it into the Balmer-Rydberg equation and find the difference between the inverses of the squared initial and final quantum numbers: \[\frac{1}{1.094 \times 10^{-6}} = 1.097 \times 10^7 \left( \frac{1}{n_{1}^2} - \frac{1}{n_{2}^2} \right)\] Now, we can simplify this equation and solve for the difference: \(\frac{1}{n_{1}^2} - \frac{1}{n_{2}^2} = \frac{9.138 \times 10^{-7}}{1.097 \times 10^7}\) \(\frac{1}{n_{1}^2} - \frac{1}{n_{2}^2} \approx 8.33 \times 10^{-14}\) Since this absorption process involves a hydrogen atom, we know that the initial and final quantum numbers are integers. Now, we need to find the closest integer values of \(n_1\) and \(n_2\) that satisfy the above condition. Through trial and error, we can find that when \(n_1 = 1\) and \(n_2 = 5\), the difference \(\frac{1}{1^2} - \frac{1}{5^2}\) equals approximately \(8.33 \times 10^{-14}\). So, the initial value of n associated with this absorption is \(n_1 = 1\), and the final value of n is \(n_2 = 5\).

Key Concepts

Electromagnetic Spectrum Regions

The electromagnetic spectrum is a broad range of waves with varying wavelengths and energies. It includes several distinct regions each with its own unique properties. These regions are defined based on the wavelength range of the electromagnetic radiation:

• Radio waves: These have the longest wavelengths ranging from about \(10^3\) to \(10^5\) nm.
• Microwaves: These waves fall between \(10^2\) and \(10^3\) nm.
• Infrared (IR): With wavelengths from \(7 \times 10^2\) to \(10^5\) nm, infrared waves are just beyond the reach of visible light.
• Visible light: This is the light detectable by the human eye, ranging from 400 to 700 nm. It consists of all the colors we can see.
• Ultraviolet (UV): These have shorter wavelengths, from 10 to 400 nm, and can cause sunburns.
• X-rays: Very short wavelengths that range from 0.01 to 10 nm, often used in medical imaging.
• Gamma rays: These have the shortest wavelengths, less than 0.01 nm, and carry the highest energies.

The light with a wavelength of 1094 nm, like the one absorbed by a hydrogen atom, falls into the infrared region. This region is often associated with heat energy and is beyond the visibility spectrum for humans.

Quantum Numbers

Quantum numbers play a crucial role in defining the properties of electrons in an atom. They are essential to understand the electron's behavior and position within the atom. Specifically, there are four quantum numbers, but the ones most relevant to spectroscopy are the principal quantum numbers.

• Principal Quantum Number \( (n) \): This number indicates the main energy level occupied by an electron. The higher the \( n \), the farther the electron is from the nucleus. In the exercise, the transition involves changes in this number, illustrating energy absorption.
• Orbital Angular Momentum Quantum Number \( (l) \): It defines the shape of the electron's orbit, but is less relevant for simple transitions.
• Magnetic Quantum Number \( (m_l) \): It specifies the orientation of the orbitals, and mostly used in complex transitions.
• Spin Quantum Number \( (m_s) \): Describes the electron's spin, important in determining the electron’s magnetic moment, but not directly involved in energy transitions.

For the hydrogen atom, the transition from \( n=1 \) to \( n=5 \) as described in the exercise represents a significant jump involving substantial energy change and absorption in the infrared spectrum.

Balmer-Rydberg Equation

The Balmer-Rydberg Equation is pivotal in calculating the wavelengths of spectral lines emitted or absorbed by hydrogen atoms. It connects the wavelength \( \lambda \) with the principal quantum numbers \( n_1 \) and \( n_2 \).The formula is given by:\[\frac{1}{\lambda} = R_H \left( \frac{1}{n_{1}^2} - \frac{1}{n_{2}^2} \right)\]Here,

• \( \lambda \): Indicates the wavelength of the light absorbed or emitted.
• \( R_H \): The Rydberg constant, approximately \(1.097 \times 10^7\) m\(^{-1}\).
• \( n_1 \) and \( n_2 \): Principal quantum numbers, where \( n_1 \) is the initial and \( n_2 \) is the final state.

In this specific instance, using the given wavelength of 1094 nm, the equation helps find that the transition involves the electron jumping from \( n = 1 \) to \( n = 5 \). This results in absorption in the infrared spectrum, emphasizing the energies associated with these quantum states.