Question

Which has higher energy, infrared radiation with \(\lambda=1.0 \times 10^{-6} \mathrm{~m}\) or an \(\mathrm{X}\) ray with \(\lambda=3.0 \times 10^{-9} \mathrm{~m}\) ? Radiation with \(\nu=4.0 \times 10^{9} \mathrm{~Hz}\) or with \(\lambda=9.0 \times 10^{-6} \mathrm{~m} ?\)

Short answer

The X-ray and the radiation with \( \lambda = 9.0 \times 10^{-6} \text{ m} \) have higher energy.

Step-by-step solution

Understand Energy and Wavelength Relationship

The energy of electromagnetic radiation is given by the equation \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( (6.63 \times 10^{-34} \text{ J}\cdot\text{s}) \), \( c \) is the speed of light \((3.00 \times 10^8 \text{ m/s}) \), and \( \lambda \) is the wavelength of the radiation.

Calculate Energy of Infrared Radiation

Given the infrared radiation has a wavelength \( \lambda = 1.0 \times 10^{-6} \text{ m} \), plug this value into the energy equation: \ \[ E_{\text{infrared}} = \frac{(6.63 \times 10^{-34}) (3.00 \times 10^8)}{1.0 \times 10^{-6}} \approx 1.989 \times 10^{-19} \text{ J} \]

Calculate Energy of X-Ray

For the \( X \) ray with \( \lambda = 3.0 \times 10^{-9} \text{ m} \): \ \[ E_{\text{x-ray}} = \frac{(6.63 \times 10^{-34}) (3.00 \times 10^8)}{3.0 \times 10^{-9}} \approx 6.63 \times 10^{-17} \text{ J} \]

Compare Energies of Infrared Radiation and X-Ray

The \( X \) ray energy \( (6.63 \times 10^{-17} \text{ J}) \) is greater than the infrared radiation energy \( (1.989 \times 10^{-19} \text{ J}) \). Therefore, the X-ray has higher energy.

Calculate Energy of Radiation with Frequency

The energy of radiation can also be calculated using \( E = h u \). For \( u = 4.0 \times 10^{9} \text{ Hz} \): \ \[ E = (6.63 \times 10^{-34}) (4.0 \times 10^{9}) = 2.652 \times 10^{-24} \text{ J} \]

Calculate Energy of Radiation with Wavelength

Given \( \lambda = 9.0 \times 10^{-6} \text{ m} \), use \( E = \frac{hc}{\lambda} \): \ \[ E = \frac{(6.63 \times 10^{-34}) (3.00 \times 10^8)}{9.0 \times 10^{-6}} \approx 2.21 \times 10^{-20} \text{ J} \]

Compare Energies Based on Frequency and Wavelength

Energy from wavelength \( (2.21 \times 10^{-20} \text{ J}) \) is greater than energy from frequency \( (2.652 \times 10^{-24} \text{ J}) \). Thus, the radiation with \( \lambda = 9.0 \times 10^{-6} \) m has higher energy.

Key concepts

Wavelength-energy relationship

The wavelength-energy relationship is a fundamental concept in understanding electromagnetic radiation. It describes the inverse relationship between the energy of radiation and its wavelength. This means that as the wavelength decreases, the energy increases, and vice versa. This concept can be expressed with the equation:\[ E = \frac{hc}{\lambda} \]Here, \(E\) is the energy of the radiation, \(h\) is Planck's constant \((6.63 \times 10^{-34} \text{ J}\cdot\text{s})\), \(c\) is the speed of light \((3.00 \times 10^8 \text{ m/s})\), and \(\lambda\) is the wavelength. So, understanding this relationship helps us predict how changing one property affects the other. It's particularly useful when comparing different types of electromagnetic waves, like infrared radiation and X-rays. For instance, shorter wavelengths, such as X-rays, have much higher energy than longer wavelengths like infrared light. This is essential for understanding why X-rays can penetrate materials that other types of radiation cannot.

Planck's constant

Planck's constant \((h)\) is a key figure in quantum mechanics and plays a crucial role in the study of electromagnetic radiation. It is a constant value used to describe the sizes of quanta or energy packets. The value of Planck's constant is approximately \(6.63 \times 10^{-34}\) joule seconds. This constant shows up in several fundamental equations in science, including the equation for energy involving radiation.

• It helps establish the quantized nature of energy.
• It is pivotal in the relationship between energy and frequency in quantum theory.
• Ensures a consistent value for calculations involving electromagnetic radiation.

In the equation \(E = \frac{hc}{\lambda}\), \(h\) denotes that each photon has a discrete energy level, contributing to phenomena like the photoelectric effect and blackbody radiation. These applications highlight the importance of Planck's constant in bridging the microscopic and macroscopic realms.

Energy calculation of radiation

Calculating the energy of electromagnetic radiation allows us to understand its capacity to perform work or enact changes at an atomic or molecular level. Energy can be derived from either wavelength or frequency, depending on the data available.Using the wavelength:- The formula \(E = \frac{hc}{\lambda}\) can be used, where larger wavelengths result in lower energy.Using frequency:- The formula \(E = h u\) is applied, where \(u\) is the frequency. Here, higher frequencies indicate greater energy due to their direct relationship with energy.Consider a practical example: when comparing different types of radiation, like infrared and X-rays, the one with the shorter wavelength (X-rays) will typically have higher energy. Similarly, radiation with higher frequency will deposit more energy per photon. Both concepts are used to solve real-world problems, such as determining energy emissions in various scientific fields, including astronomy and medical imaging applications. It's crucial for detailed understanding of how radiation interacts with matter, affecting everything from chemical reactions to diagnostic imaging in healthcare.