Question

Rank these photons in terms of decreasing energy: IR ( \(\nu=\) \(\left.6.5 \times 10^{13} \mathrm{~s}^{-1}\right) ;\) microwave \(\left(\nu=9.8 \times 10^{11} \mathrm{~s}^{-1}\right) ; \mathrm{UV}\left(\nu=8.0 \times 10^{15} \mathrm{~s}^{-1}\right)\)

Short answer

UV, IR, Microwave

Step-by-step solution

- Recall the relationship between frequency and energy

Photon energy can be calculated using the equation: \[ E = h u \]where \( E \) is the energy of the photon, \( h \) is Planck's constant (\(6.626 \times 10^{-34}~ \text{Js} \)), and \( u \) is the frequency of the photon.

- Plug in the frequency values

Given the frequencies:IR: \( u = 6.5 \times 10^{13}~ \text{s}^{-1} \)Microwave: \( u = 9.8 \times 10^{11}~ \text{s}^{-1} \)UV: \( u = 8.0 \times 10^{15}~ \text{s}^{-1} \)

- Determine the energy for each type of photon

Calculate the energy for each type of photon using the equation from Step 1:IR: \[ E_{IR} = h \times 6.5 \times 10^{13} = 6.626 \times 10^{-34}~ \text{Js} \times 6.5 \times 10^{13}~ \text{s}^{-1} = 4.307 \times 10^{-20}~ \text{J} \]Microwave: \[ E_{microwave} = h \times 9.8 \times 10^{11} = 6.626 \times 10^{-34}~ \text{Js} \times 9.8 \times 10^{11}~ \text{s}^{-1} = 6.493 \times 10^{-22}~ \text{J} \]UV: \[ E_{UV} = h \times 8.0 \times 10^{15} = 6.626 \times 10^{-34}~ \text{Js} \times 8.0 \times 10^{15}~ \text{s}^{-1} = 5.301 \times 10^{-18}~ \text{J} \]

- Rank the photon energies in decreasing order

Compare the calculated energies:\[ E_{UV} = 5.301 \times 10^{-18}~ \text{J} \]\[ E_{IR} = 4.307 \times 10^{-20}~ \text{J} \]\[ E_{microwave} = 6.493 \times 10^{-22}~ \text{J} \]Thus, the ranking in decreasing order of energy is UV, IR, Microwave.

Key concepts

Frequency-Energy Relationship

In physics, the energy of a photon is directly related to its frequency. This means that the higher the frequency of the photon, the higher its energy will be. This relationship is expressed using the equation: \( E = hu \). Here, \( E \) represents the energy of the photon, \( h \) is Planck's constant, and \( u \) (nu) is the frequency of the photon. This equation shows that energy is proportional to frequency. For a clear example, if you compare different types of radiation, like infrared (IR), microwaves, and ultraviolet (UV), you’ll see that photons with higher frequencies (like UV light) have more energy than those with lower frequencies (like microwaves and IR).

Planck's Constant

Planck’s constant (\( h \)) is a fundamental constant in physics that plays a crucial role in the quantum mechanics of particles. Its value is approximately \( 6.626 \times 10^{-34} \text{Js} \). Planck's constant is essential in the equation \( E = h u \), which connects the energy of a photon with its frequency. This constant signifies the smallest possible unit of energy transfer. To understand it with an example, consider the universal equation for photon energy: \( E = h u \). By knowing the value of Planck's constant and the frequency of a photon, you can calculate its energy. For example, if you have a photon with a frequency of \( 6.5 \times 10^{13} \text{s}^{-1} \) (as in infrared radiation), multiplying this frequency by Planck's constant gives the photon's energy. This makes Planck’s constant a vital element in understanding and calculating photon energies in various situations.

Electromagnetic Spectrum

The Electromagnetic Spectrum encompasses all types of electromagnetic radiation, which vary in wavelength and frequency. It ranges from long-wavelength radio waves to short-wavelength gamma rays. Different parts of the spectrum include:

• Radio Waves
• Microwaves
• Infrared (IR)
• Visible Light
• Ultraviolet (UV)
• X-rays
• Gamma Rays

Each type of radiation within the spectrum has its own range of wavelengths and frequencies. For example, radio waves have long wavelengths and low frequencies, while gamma rays have very short wavelengths and very high frequencies. Understanding the electromagnetic spectrum helps in identifying and ranking the energy of different types of photons based on their frequencies. Photons with higher frequencies, such as UV rays, contain more energy compared to lower frequency photons like microwaves or IR rays. This is why UV radiation is more energetic and can cause more damage to living tissues compared to IR or microwaves.