Question

Without doing detailed calculations, indicate which of the following electromagnetic radiations has the greatest energy per photon and which has the least: (a) \(662 \mathrm{nm}\) (b) \(2.1 \times 10^{-5} \mathrm{cm} ;\) (c) \(3.58 \mu \mathrm{m} ;\) (d) \(4.1 \times 10^{-6} \mathrm{m}\).

Short answer

Photon (d) with wavelength \(4.1 \times 10^{-6} m\) has the greatest energy and photon (b) with wavelength \(2.1 \times 10^{-7} m\) has the least energy.

Step-by-step solution

Transform Wavelength Units

First, convert all wavelengths to the same unit for a proper comparison. The comprehensive unit will be meter (m) - the SI unit for length. The conversions are: (a) \(662 nm = 662 \times 10^{-9} m\), (b) \(2.1 \times 10^{-5} cm = 2.1 \times 10^{-7} m\), (c) \(3.58 \mu m = 3.58 \times 10^{-6} m\), (d) \(4.1 \times 10^{-6} m\) remains the same.

Identify the Photon with Greatest Energy

To identify the photon with the greatest energy, look for the radiation with the shortest wavelength. As calculated in the previous step, (d) \(4.1 \times 10^{-6} m\) is the shortest wavelength.

Identify the Photon with Least Energy

To identify the photon with the least energy, look for the radiation with the longest wavelength. As calculated in step 1, (b) \(2.1 \times 10^{-7} m\) is the longest wavelength.

Key concepts

Photon Energy

Photon energy is a fundamental concept in understanding electromagnetic radiation. Every photon, a quantum of electromagnetic energy, carries energy that is inversely proportional to its wavelength. This relationship can be expressed through the equation: \[ E = \frac{hc}{\lambda} \]where:- \( E \) is the photon energy,- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \text{ m}^2 \cdot \text{kg/s}\)),- \( c \) is the speed of light in a vacuum (\(3.00 \times 10^8 \text{ m/s}\)),- \( \lambda \) is the wavelength of the radiation.This means that shorter wavelengths yield higher energy photons, while longer wavelengths correspond to lower energy photons. Grasping this relationship is crucial for analyzing the energy of different types of electromagnetic radiation. Whether it be analyzing visible light, ultraviolet, or infrared radiation, understanding photon energy allows us to explore how light interacts with matter.

Wavelength Conversion

When comparing different electromagnetic radiations, it's often necessary to convert their wavelengths into a consistent unit. This ensures that accurate comparisons can be made.The standard unit of length in the International System (SI) is the meter (m). Converting different wavelength measurements to meters can involve the following:

• For nanometers (nm): Use the conversion \(1 \, nm = 1 \, \times 10^{-9} \, m\).
• For micrometers (μm): Use the conversion \(1 \, \mu m = 1 \, \times 10^{-6} \, m\).
• For centimeters (cm): Use the conversion \(1 \, cm = 1 \, \times 10^{-2} \, m\).

By converting all wavelengths to meters, we can easily compare their values to determine which has the greater or lesser photon energy. For the conversion, you may multiply the original value by the power of ten that corresponds to its respective conversion factor. Maintaining consistency in units is key to solving problems related to electromagnetic radiation efficiently.

Wavelength and Energy Relationship

The relationship between the wavelength of electromagnetic radiation and its energy is crucial to understanding its behavior and properties. This relationship is primarily determined by the formula stating that energy is inversely proportional to wavelength:\[ E \propto \frac{1}{\lambda} \]This inverse relationship implies several important principles:

• Shorter wavelengths, such as those in the ultraviolet spectrum, are associated with higher energy photons. These wavelengths can have significant impacts, such as causing sunburn.
• Longer wavelengths, like those in the infrared and radio wave spectrums, correspond to lower energy photons, making them less intense in terms of energy transfer.

By understanding this relationship, you can identify high-energy or low-energy photons by simply inspecting their wavelengths. This principle is applicable across many scientific fields, from quantum mechanics to medical imaging and telecommunications, where different types of electromagnetic radiations are employed based on the energy they carry.