Problem 125 - Chemistry Matter and Change

Question

A mercury atom drops in energy from \(1.413 \times 10^{-18} \mathrm{J}\) to \(1.069 \times 10^{-18} \mathrm{J.(Chapter} 5 )\) a. What is the energy of the photon emitted by the mercury atom? b. What is the frequency of the photon emitted by the mercury atom? c. What is the wavelength of the photon emitted by the mercury atom?

Step-by-Step Solution

Answer

The photon emitted has an energy of \(0.344 \times 10^{-18} \text{ J}\), a frequency of \(5.19 \times 10^{14} \text{ Hz}\), and a wavelength of \(5.78 \times 10^{-7} \text{ m}\).

Step 1: Calculate Energy of the Emitted Photon

The energy of the emitted photon is equal to the change in energy of the mercury atom. We calculate this by subtracting the final energy from the initial energy.\[E = E_{initial} - E_{final} \]Given:\[ E_{initial} = 1.413 \times 10^{-18} \text{ J} \]\[ E_{final} = 1.069 \times 10^{-18} \text{ J} \]Thus,\[E = 1.413 \times 10^{-18} \text{ J} - 1.069 \times 10^{-18} \text{ J} = 0.344 \times 10^{-18} \text{ J}\]

Step 2: Calculate Frequency of the Photon

The frequency of the photon can be found using the formula for energy in terms of frequency:\[E = h \cdot f \]Where:- \(E\) is the energy of the photon,- \(h = 6.626 \times 10^{-34} \text{ Js}\) is Planck's constant,- \(f\) is the frequency of the photon.Rearrange the formula to solve for \(f\):\[f = \frac{E}{h} \]Substitute the known values:\[f = \frac{0.344 \times 10^{-18} \text{ J}}{6.626 \times 10^{-34} \text{ Js}} \approx 5.19 \times 10^{14} \text{ Hz}\]

Step 3: Calculate Wavelength of the Photon

The wavelength of the photon can be calculated using the wave equation that relates the speed of light, frequency, and wavelength:\[c = f \cdot \lambda \]Where:- \(c = 3.00 \times 10^{8} \text{ m/s}\) is the speed of light,- \(f\) is the frequency,- \(\lambda\) is the wavelength.Rearrange the formula to solve for \(\lambda\):\[\lambda = \frac{c}{f} \]Substitute the known values:\[\lambda = \frac{3.00 \times 10^{8} \text{ m/s}}{5.19 \times 10^{14} \text{ Hz}} \approx 5.78 \times 10^{-7} \text{ m} \]

Key Concepts

Energy Levels

When atoms change their energy state, they either absorb or emit energy in the form of photons. This process is closely associated with the concept of energy levels. Imagine energy levels like the rungs of a ladder; electrons can "jump" from one level to another when an atom gains or loses energy.

For the mercury atom, we are dealing with a drop in energy, which means that it releases energy. This release is quantified by the difference in the atom's initial and final energy states. The formula used is:

• Initial energy: \( E_{initial} = 1.413 \times 10^{-18} \text{ J} \)
• Final energy: \( E_{final} = 1.069 \times 10^{-18} \text{ J} \)
• Energy of the emitted photon: \( E = E_{initial} - E_{final} = 0.344 \times 10^{-18} \text{ J} \)

By calculating this difference, it's easy to determine the energy of the emitted photon. This photon is the result of the electron transitioning from a higher to a lower energy level.

Frequency Calculation

Understanding the concept of frequency is crucial when dealing with photons. Frequency refers to how many times a wave, such as those found in light, passes a point in one second.

The energy of a photon is directly related to its frequency by Planck's equation:

• Planck's equation: \( E = h \cdot f \)
• where \( h = 6.626 \times 10^{-34} \text{ Js} \) is Planck's constant
• \( f \) is the frequency.

Rearranging for frequency, we get:

• \( f = \frac{E}{h} \)
• Substitute the values:\( f = \frac{0.344 \times 10^{-18} \text{ J}}{6.626 \times 10^{-34} \text{ Js}} \approx 5.19 \times 10^{14} \text{ Hz} \)

Frequency is reported in Hertz (Hz), which tells us the number of times the wave oscillates per second. Here, the frequency of the emitted photon is approximately \( 5.19 \times 10^{14} \text{ Hz} \), highlighting how rapid these oscillations are.

Wavelength Calculation

Wavelength is another essential property of photons, offering a different perspective than frequency. While frequency addresses how often waves occur in a time period, wavelength focuses on the distance between two consecutive peaks of the wave.

For photons, the relationship between wavelength \(\lambda\) and frequency \(f\) is described by the speed of light formula:

• \( c = f \cdot \lambda \)
• where \( c = 3.00 \times 10^{8} \text{ m/s} \) is the speed of light.

Rearranging to solve for wavelength:

• \( \lambda = \frac{c}{f} \)
• Substitute the known frequency:\( \lambda = \frac{3.00 \times 10^{8} \text{ m/s}}{5.19 \times 10^{14} \text{ Hz}} \approx 5.78 \times 10^{-7} \text{ m} \)

The wavelength found here indicates the space between wave peaks as about \( 5.78 \times 10^{-7} \text{ m} \), showcasing how close together the waves are in the electromagnetic spectrum. This wavelength typically falls within the visible range, explaining the light emission we see during such photon emissions.