Question

A solution of KMnO absorbs light at 540 nm (page 192). What is the frequency of the light absorbed? What is the energy of one mole of photons with \(\lambda=540 \mathrm{nm} ?\)

Short answer

Frequency: \(5.56 \times 10^{14} \text{ Hz}\); Energy/mole: \(2.22 \times 10^{5} \text{ J/mol}\).

Step-by-step solution

Identify Known Variables

We know the wavelength of the absorbed light, \( \lambda = 540 \text{ nm} \). We need to convert this to meters for proper units in calculations: \( 540 \text{ nm} = 540 \times 10^{-9} \text{ m} \).

Calculate the Frequency

Use the speed of light equation \( c = \lambda \cdot f \) to find the frequency \( f \). Here, \( c \) is the speed of light \( 3 \times 10^8 \text{ m/s} \), and \( \lambda \) is \( 540 \times 10^{-9} \text{ m} \). Rearrange the equation to solve for \( f \):\[ f = \frac{c}{\lambda} = \frac{3 \times 10^8 \text{ m/s}}{540 \times 10^{-9} \text{ m}} = 5.56 \times 10^{14} \text{ Hz} \]

Calculate Energy of One Photon

Use Planck’s equation \( E = h \cdot f \) to calculate the energy of one photon, where \( h \) is Planck’s constant \( 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \). Substitute the frequency calculated in Step 2:\[ E = 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \times 5.56 \times 10^{14} \text{ Hz} = 3.68 \times 10^{-19} \text{ J} \]

Calculate Energy of One Mole of Photons

For energy per mole, use Avogadro’s number \( 6.022 \times 10^{23} \text{ mol}^{-1} \). Calculate the energy of one mole of photons by multiplying the energy of one photon by Avogadro’s number:\[ E_{\text{mole}} = 3.68 \times 10^{-19} \text{ J} \times 6.022 \times 10^{23} \text{ mol}^{-1} = 2.22 \times 10^{5} \text{ J/mol} \]

Key concepts

Photon Energy

When discussing photon energy, we look at the small packets of light energy that photons carry. This energy is quantified using Planck’s equation:

• \( E = h \cdot f \)
• Here, \( E \) represents the energy of a single photon.
• \( h \) is Planck’s constant, valued at \( 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \).
• \( f \) is the frequency of the light, which denotes how many waves pass a point per second.

Understanding photon energy helps us connect the concepts of light with energy, which is crucial in chemical reactions and in understanding molecular energy transitions. For example, the calculated photon energy of \( 3.68 \times 10^{-19} \text{ J} \) from the given solution is the amount of energy each photon carries at the specified frequency. Knowing this helps chemists determine how light interacts with materials on a molecular level. Whether in photosynthesis or semiconductors, photon energy plays a critical role.

Frequency Calculation

Frequency calculation is central to determining the characteristics of electromagnetic waves. In our problem, we use the relationship between speed of light \( c \), wavelength \( \lambda \), and frequency \( f \):

• \( c = \lambda \cdot f \)
• The speed of light \( c \) is a constant \( 3 \times 10^8 \text{ m/s} \).
• To find the frequency \( f \), rearrange the equation: \( f = \frac{c}{\lambda} \).

The light’s wavelength is given in nanometers, so it’s important to convert this to meters. For 540 nm, we have:

• \( 540 \text{ nm} = 540 \times 10^{-9} \text{ m} \)
• Then, \( f = \frac{3 \times 10^8 \text{ m/s}}{540 \times 10^{-9} \text{ m}} = 5.56 \times 10^{14} \text{ Hz} \).

Frequency is crucial as it determines the photon’s energy and can vary across the electromagnetic spectrum, affecting visibility and other properties of light.

Avogadro's Number

Avogadro's Number is a fundamental constant in chemistry linking the atomic scale to macroscale quantities. It defines the number of entities (such as atoms or molecules) in a mole:

• Avogadro’s number is \( 6.022 \times 10^{23} \text{ mol}^{-1} \).
• This helps convert the energy of a single photon to the energy of a whole mole of photons.
• In our problem, once we know the energy per photon, we can calculate the energy for a mole of photons by multiplying \( 3.68 \times 10^{-19} \text{ J} \) by Avogadro's number.
• The result, \( 2.22 \times 10^{5} \text{ J/mol} \), shows the large-scale energy impact due to the microscopic photon energy.

This constant is pivotal in bridging scales and is indispensable in calculations involving substances at the molecular level, affecting reactions and energy changes in chemistry and physics alike.