Problem 82 - Chemistry: The Central Science in SI Units, Global Edition

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

The frequency of the emitted light is \(7.10 * 10^{14} Hz\). The energy of 1 mole of photons is \(2.84 * 10^{5} J/mol\). The energy gap between the excited and ground states for the calcium atom is \(4.71 * 10^{-19} J\).

Step 1: Convert wavelength to meters

We are given the wavelength in nanometers (nm), so we need to convert it to meters (m) for our calculations. Use the conversion factor 1 nm = 10⁻⁹ m: λ = 422.7 nm * (1 * 10⁻⁹ m / 1 nm) = 422.7 * 10⁻⁹ m

Step 2: Calculate the frequency

Now that we have the wavelength in meters, we can use the speed of light equation to calculate the frequency: c = λ * ν Rearrange the equation to solve for frequency: ν = c / λ ν = (3 * 10⁸ m/s) / (422.7 * 10⁻⁹ m) = 7.10 * 10¹⁴ Hz The frequency of the emitted light is 7.10 * 10¹⁴ Hz. #b) Calculating the energy of 1 mole of photons#

Step 3: Calculate the energy of one photon

Now that we have the frequency, we can use Planck's equation to calculate the energy of one photon: E = h * ν E = (6.626 * 10⁻³⁴ Js) * (7.10 * 10¹⁴ Hz) = 4.71 * 10⁻¹⁹ J The energy of one photon is 4.71 * 10⁻¹⁹ J.

Step 4: Calculate the energy of 1 mole of photons

To find the energy of 1 mole of photons, multiply the energy of one photon by Avogadro's number (Nₐ): E₁ (1 mole of photons) = Nₐ * E E₁ = (6.022 * 10²³ mol⁻¹) * (4.71 * 10⁻¹⁹ J) = 2.84 * 10⁵ J/mol The energy of 1 mole of photons is 2.84 * 10⁵ J/mol. #c) Calculating the energy gap between the excited and ground states of the calcium atom# Since the energy of a single photon corresponds to the energy gap between the excited and ground states of the calcium atom, we have already found the required energy gap in step 3. Thus, the energy gap between the excited and ground states for the calcium atom is 4.71 * 10⁻¹⁹ J.

Key Concepts

Wavelength Calculation

Wavelength is an essential concept in understanding the behavior of light. When calcium chloride is introduced into a nonluminous flame, the wavelength of the emitted orange light can be measured. In this exercise, the wavelength is given as \(422.7\, \mathrm{nm}\). To use this in calculations, it's essential to convert it from nanometers to meters because the standard unit in physics for wavelength is meters. To convert nanometers to meters, use the conversion factor:

• 1 nanometer = \(1 \times 10^{-9}\) meters.

This means:\[ \lambda = 422.7 \, \mathrm{nm} \times \left(1 \times 10^{-9} \, \mathrm{m/nm}\right) = 422.7 \times 10^{-9} \, \mathrm{m} \]Converting to meters makes it straightforward to plug into other formulas, like calculating frequency.

Frequency Calculation

Frequency is a measure of how many wavelengths pass a given point per unit of time. It is denoted by the Greek letter nu (\(u\)). To find the frequency of light emitted from the excited calcium, we use the speed of light equation: \[ c = \lambda \times u \]Where:

• \( c \) is the speed of light \(\left(3 \times 10^8 \, \mathrm{m/s} \right)\).
• \( \lambda \) is the wavelength in meters \(\left(422.7 \times 10^{-9} \, \mathrm{m}\right)\).

Rearrange the equation to solve for frequency:\[ u = \frac{c}{\lambda} \]Substitute in the values:\[ u = \frac{3 \times 10^8 \, \mathrm{m/s}}{422.7 \times 10^{-9} \, \mathrm{m}} \approx 7.10 \times 10^{14} \, \mathrm{Hz} \]This means that the frequency of the emitted light is \(7.10 \times 10^{14}\) Hertz, indicating how rapidly the light waves are oscillating.

Photon Energy

The energy of a photon relates directly to its frequency. We use Planck's equation to find the energy associated with one photon of calcium's light emission:\[ E = h \times u \] Where:

• \( E \) is the energy of the photon.
• \( h \) is Planck's constant \(\left(6.626 \times 10^{-34} \, \mathrm{Js}\right)\).
• \( u \) is the frequency \(\left(7.10 \times 10^{14} \, \mathrm{Hz}\right)\).

Substitute in these values:\[ E = 6.626 \times 10^{-34} \, \mathrm{Js} \times 7.10 \times 10^{14} \, \mathrm{Hz} = 4.71 \times 10^{-19} \, \mathrm{J} \]This result gives the energy for an individual photon. To determine the energy for a mole of these photons, multiply by Avogadro's number (\(6.022 \times 10^{23} /\mathrm{mol}\)):\[ E_1 = 6.022 \times 10^{23} \times 4.71 \times 10^{-19} \, \mathrm{J} = 2.84 \times 10^5 \, \mathrm{J/mol} \]This value signifies the collective energy carried by one mole of these light particles, termed an Einstein.

Chemical Excitation

Chemical excitation occurs when electrons in atoms absorb energy and move to a higher energy level. This process is temporary, and when these electrons transition back to their original states, they release energy in the form of light. For calcium in a flame test, the atoms become excited by absorbing thermal energy. When returning to the ground state, the orange light is emitted. This emission corresponds to the energy gap between the excited and ground states. In practical terms, for the calcium atom:

• The energy difference (or gap) given by the photon's energy in this process is \(4.71 \times 10^{-19} \, \mathrm{J}\).

The distinct color emitted helps us identify specific elements, showcasing the significance in fields such as analytical chemistry and astrophysics.