Infrared, visible and ultraviolet radiations /light have different properties. Which one of the following statements related to these radiations/lights is not correct?
The photon energy of ultraviolet is lesser than that of visible light.
Understanding Electromagnetic Radiation Properties
Electromagnetic radiation, including infrared, visible light, and ultraviolet radiation, travels in waves and carries energy. These types of radiation are part of the electromagnetic spectrum, differing primarily in their wavelength and frequency. A key relationship connects these properties to the energy carried by the radiation's photons.
The energy of a photon (\(E\)) is directly proportional to its frequency (\(\nu\)) and inversely proportional to its wavelength (\(\lambda\)). This relationship is given by the equation:
\[E = h\nu = \frac{hc}{\lambda}\]
where \(h\) is Planck's constant and \(c\) is the speed of light in a vacuum.
Let's consider the order of these radiations on the electromagnetic spectrum in terms of increasing frequency and energy, and decreasing wavelength:
Infrared (\(\rightarrow\)) Visible Light (\(\rightarrow\)) Ultraviolet
This means:
- Wavelength: Infrared > Visible Light > Ultraviolet
- Frequency: Infrared < Visible Light < Ultraviolet
- Photon Energy: Infrared < Visible Light < Ultraviolet
Analyzing Statements about Radiation Properties
Now let's evaluate each given statement based on this understanding of the electromagnetic spectrum:
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Statement 1: The wavelength of infrared is more than that of ultraviolet radiation.
Based on the spectrum order, infrared radiation has a longer wavelength than visible light, and visible light has a longer wavelength than ultraviolet radiation. Therefore, infrared radiation has a significantly longer wavelength than ultraviolet radiation. This statement is correct.
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Statement 2: The wavelength of ultraviolet is smaller than that of visible light.
Moving across the spectrum from infrared towards ultraviolet, the wavelength decreases. Visible light is located before ultraviolet radiation. Therefore, ultraviolet radiation has a shorter wavelength than visible light. This statement is correct.
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Statement 3: The photon energy of visible light is more than that of infrared light.
Energy is inversely proportional to wavelength and directly proportional to frequency. As we move from infrared to visible light, the frequency increases and the wavelength decreases, leading to higher photon energy. Therefore, visible light photons have more energy than infrared light photons. This statement is correct.
-
Statement 4: The photon energy of ultraviolet is lesser than that of visible light.
Moving from visible light to ultraviolet radiation, the frequency increases and the wavelength decreases. This corresponds to an increase in photon energy. Therefore, ultraviolet radiation photons have more energy than visible light photons, not less. This statement is incorrect.
The statement that is not correct is "The photon energy of ultraviolet is lesser than that of visible light."
Summary Comparison of Radiation Properties
| Property | Infrared (IR) | Visible Light (Vis) | Ultraviolet (UV) |
|---|---|---|---|
| Wavelength (\(\lambda\)) | Longest | Medium | Shortest |
| Frequency (\(\nu\)) | Lowest | Medium | Highest |
| Photon Energy (E) | Lowest | Medium | Highest |
Revision Table: Electromagnetic Spectrum
Understanding the order and properties of different types of electromagnetic radiation is crucial. The spectrum ranges from low-energy, long-wavelength radio waves to high-energy, short-wavelength gamma rays. Infrared, visible, and ultraviolet radiation are specific regions within this broad spectrum.
- Radio Waves
- Microwaves
- Infrared
- Visible Light
- Ultraviolet
- X-rays
- Gamma Rays
Wavelength decreases and frequency/energy increases as you move from left to right in this list.
Additional Information: Photon Energy and Wavelength
The inverse relationship between photon energy and wavelength (\(E = hc/\lambda\)) means that shorter wavelengths correspond to higher energy photons, and longer wavelengths correspond to lower energy photons. This principle explains why ultraviolet radiation, despite having less power in some applications compared to visible light sources, consists of photons that are individually more energetic. This higher individual photon energy is why UV radiation can cause more damage at the molecular level (like sunburn) compared to visible light.
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