Problem 82 - Chemistry The Central Science In Si Units

Question

If a sample of calcium chloride is introduced into a nonluminous flame, the color of the flame turns to orange ("flame test"). The light is emitted because calcium atoms become excited; their return to the ground state results in light emission. (a) The wavelength of this emitted light is \(422.7 \mathrm{nm} .\) Calculate its frequency. (b) What is the energy of \(1.00 \mathrm{~mol}\) of these photons (a mole of photons is called an Einstein)? (c) Calculate the energy gap between the excited and ground states for the calcium atom.

Step-by-Step Solution

Answer

(a) Frequency: \( 7.10 \times 10^{14} \) Hz. (b) Energy of 1 mole: \( 2.83 \times 10^{5} \) J/mol. (c) Energy gap: \( 4.70 \times 10^{-19} \) J per photon.

Step 1: Understanding Wavelength and Frequency

The frequency \( f \) of light is related to its wavelength \( \lambda \) through the speed of light \( c \). The formula is: \[ f = \frac{c}{\lambda} \]where \( c \approx 3.00 \times 10^8 \text{ m/s} \) and \( \lambda = 422.7 \text{ nm} = 422.7 \times 10^{-9} \text{ m} \).

Step 2: Calculating Frequency

Substitute the values into the formula:\[ f = \frac{3.00 \times 10^8 \text{ m/s}}{422.7 \times 10^{-9} \text{ m}} \approx 7.10 \times 10^{14} \text{ Hz} \]

Step 3: Understanding Energy of Photons

The energy \( E \) of a photon is given by\[ E = hf \]where \( h = 6.626 \times 10^{-34} \text{ J·s} \) is Planck's constant, and \( f \) is the frequency calculated in Step 2.

Step 4: Calculating Energy of One Photon

Using the frequency from Step 2:\[ E = (6.626 \times 10^{-34} \text{ J·s})(7.10 \times 10^{14} \text{ Hz}) \approx 4.70 \times 10^{-19} \text{ J} \]

Step 5: Calculating Energy of 1 Mole of Photons

Multiply the energy of one photon by Avogadro's number \( N_A \approx 6.022 \times 10^{23} \text{ mol}^{-1} \):\[ E_{\text{mole}} = (4.70 \times 10^{-19} \text{ J/photon}) (6.022 \times 10^{23} \text{ photons/mol}) \approx 2.83 \times 10^5 \text{ J/mol} \]

Step 6: Understanding Energy Gap

The energy gap between the excited and ground states of the calcium atom corresponds to the energy of the photons emitted, calculated in Step 4.

Key Concepts

Flame Test

The flame test is a simple method used to identify certain metallic ions based on the characteristic colors they emit when heated in a flame. The test relies on the behavior of electrons in an atom. When a metal ion is introduced into a nonluminous flame, the heat provides energy to the electrons, causing them to jump to higher energy levels, or "excited states."
When these electrons return to their original energy levels, or "ground states," they release energy in the form of light. This light often appears as a distinct color that can help identify the metal. For example, calcium ions emit an orange color, which is characteristic of this element.
It's important to note that each element has a unique emission spectrum, meaning the color of light emitted can act like a fingerprint. The flame test is a straightforward tool in labs to quickly screen for the presence of specific metal ions in a sample without using complex equipment.

Frequency of Light

The frequency of light is a fundamental property that describes the number of wave cycles that pass a particular point in one second. It is measured in units of Hertz (Hz). Understanding frequency helps in studying various aspects of light and electromagnetic waves.
Frequency is inversely related to wavelength, the distance between successive peaks of a wave, and directly proportional to the speed of light. In general terms, a shorter wavelength means a higher frequency, and vice versa. This relationship is expressed by the formula:

• \[ f = \frac{c}{\lambda} \]

where

• \( f \) is the frequency
• \( c \) is the speed of light (approximately \( 3.00 \times 10^8 \text{ m/s} \))
• \( \lambda \) is the wavelength

To calculate the frequency of light with a known wavelength:

• Convert the wavelength into meters if it is not already.
• Use the formula provided to find \( f \).

This calculation is critical in photon energy calculations because it is directly used to find the energy of a photon.

Wavelength

Wavelength is a key concept in understanding the nature of light. It is defined as the distance between successive peaks of a wave. Different wavelengths of light are perceived as different colors, which is why this property is important in many scientific fields, especially physics and chemistry.
In the context of the flame test, observing the wavelength of light emission is a way to identify the element in question. The electromagnetic spectrum covers a wide range of wavelengths, from shorter wavelengths like gamma rays, to longer wavelengths like radio waves. Visible light falls within a small portion of this spectrum.
By knowing the wavelength, one can derive other important properties of light, such as frequency and energy. The formula connecting wavelength \( \lambda \) to frequency \( f \) is:

• \[ \lambda = \frac{c}{f} \]

where

• \( c \) is the speed of light
• \( \lambda \) is the wavelength

Understanding wavelength is essential in interpreting various phenomena, from basic laboratory tests like the flame test to advanced applications in spectroscopy.

Planck's Constant

Planck's Constant is a fundamental constant in quantum mechanics, denoted by \( h \), with a value of approximately \( 6.626 \times 10^{-34} \text{ J·s} \). This constant is crucial in calculations that involve particle-like properties of electromagnetic waves.
In the study of light, Planck's constant is used to relate the energy \( E \) of a photon to its frequency \( f \) using Planck-Einstein relation:

• \[ E = hf \]

where

• \( E \) is the energy of the photon
• \( h \) is Planck's constant
• \( f \) is the frequency

This relationship highlights that energy and frequency are directly proportional. Thus, higher frequency waves carry more energy. Planck's constant plays a vital role in quantifying energy levels in atoms and is a key part of understanding atomic energy transitions. In the context of the flame test, it helps calculate the precise energy released when electrons return to lower energy states and emit light.