Question

(a) What is the frequency of radiation whose wavelength is \(10.0 \mathrm{~A}\) ? (b) What is the wavelength of radiation that has a frequency of \(7.6 \times 10^{10} \mathrm{~s}^{-1} ?\) (c) Would the radiations in part (a) or part (b) be detected by an \(\mathrm{X}\) -ray detector? (d) What distance does electromagnetic radiation travel in \(25.5 \mathrm{fs}\) ?

Short answer

The frequency of the radiation with a wavelength of \(10.0 \, Å\) is \(3.0 \times 10^{17} \, s^{-1}\), the wavelength of the radiation with a frequency of \(7.6 \times 10^{10} \, s^{-1}\) is \(3.95 \times 10^{-3} \, m\), the radiation from part (a) can be detected by an X-ray detector, but the radiation from part (b) cannot. The distance traveled by electromagnetic radiation in \(25.5 \, fs\) is \(7.65 \times 10^{-6} \, m\).

Step-by-step solution

Recall the relationship between speed of light, frequency, and wavelength

We can find the frequency and wavelength using the formula: \(c = \lambda \nu\), where: - \(c\) is the speed of light (\(3.0 \times 10^8 \, m/s\)) - \(\lambda\) is the wavelength - \(\nu\) is the frequency

Solve for frequency in part (a)

We're given the wavelength (\(\lambda = 10.0 \, Å\)) and we need to find the frequency. First, convert the given wavelength to meters: \(10.0 \, Å = 1.0 \times 10^{-9} \, m\). Now, we can solve for the frequency using the formula: \(\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \, m/s}{1.0 \times 10^{-9} \, m} = 3.0 \times 10^{17} \, s^{-1}\)

Solve for wavelength in part (b)

We're given the frequency (\(\nu = 7.6 \times 10^{10} \, s^{-1}\)) and we need to find the wavelength. Using the formula, we can solve for the wavelength: \(\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^8 \, m/s}{7.6 \times 10^{10} \, s^{-1}} = 3.95 \times 10^{-3} \, m\)

Determine if radiation from part (a) and (b) can be detected by an X-ray detector in part (c)

The wavelength range for X-rays is typically between \(0.01 \, Å\) to \(10 \, Å\). Comparing our calculated values with this range: - The radiation from part (a) has a wavelength of \(10.0 \, Å\), which falls within the X-ray range. So, it can be detected by an X-ray detector. - The radiation from part (b) has a wavelength of \(3.95 \times 10^{-3} \, m\), which is outside the X-ray range. So, it cannot be detected by an X-ray detector.

Calculate the distance traveled by electromagnetic radiation in 25.5 fs in part (d)

We're given the time (\(t = 25.5 \, fs\)) and we need to find the distance traveled by electromagnetic radiation. First, convert the given time to seconds: \(25.5 \, fs = 2.55 \times 10^{-14} \, s\). The speed of light is \(c = 3.0 \times 10^8 \, m/s\). Thus, the distance can be calculated as: Distance = \(c \times t = (3.0 \times 10^8 \, m/s) \times (2.55 \times 10^{-14} \, s) = 7.65 \times 10^{-6} \, m\)

Key concepts

Wavelength and Frequency Relationship

Understanding the relationship between wavelength and frequency is fundamental in the study of electromagnetic radiation. The formula that connects these two properties is given by the speed of light equation: \( c = \lambda u \). Here, \( \lambda \) represents the wavelength of the wave, while \( u \) stands for the frequency. Since the speed of light (\( c \) ) is a constant at approximately \( 3.0 \times 10^8 \, m/s \), when the wavelength increases, the frequency decreases and vice versa. This is an inverse relationship, which is key in solving problems related to electromagnetic waves.

This formula is essential when you're trying to find unknown values for frequency or wavelength as shown in the textbook exercise. For instance, if you have a wavelength of \( 10.0 \, Å \), which is \( 1.0 \times 10^{-9} \, m \) when converted to meters, and need to find the corresponding frequency, the rearranged formula \( u = \frac{c}{\lambda} \) allows you to calculate that. This shows how interdependent wavelength and frequency are, which is a principle cornerstone in the fields ranging from astronomy to telecommunications.

Speed of Light

The speed of light (\( c \)) is a fundamental constant that is crucially important in the field of physics and beyond. It is the speed at which all electromagnetic radiation propagates in a vacuum and is approximately \( 3.0 \times 10^8 \, m/s \). Because of its constancy, the speed of light is used as a base measurement in calculations involving time and distance, like the one outlined in the exercise where electromagnetic radiation travels a certain distance in a short amount of time.

This constant not only defines the fast pace at which light travels but also helps in determining vast distances in the cosmos, as well as in understanding concepts such as time dilation in the theory of relativity. It is the ultimate speed limit of the universe, and no known matter can travel as fast. In practical terms, when a problem involves time, like the 25.5 femtoseconds (fs) mentioned in part (d) of the exercise, it becomes solvable by multiplying this time with the speed of light to find the distance covered by radiation.

X-ray Detection

X-ray detection is a process of identifying the presence of X-rays, which are a type of electromagnetic radiation with certain wavelengths and frequencies. X-rays typically have wavelengths ranging from \( 0.01 \, Å \) to \( 10 \, Å \). As part of the exercise, it's important to know whether radiation can be detected by an X-ray detector, which hinges on whether the wavelength of the radiation falls within this range.

An X-ray detector is designed to be sensitive to this specific range and will not identify waves with significantly longer wavelengths, such as radio waves, or shorter wavelengths, like gamma rays. In the educational setting, exercises often ask about whether certain wavelengths would register on an X-ray detector to test a student's comprehension of the electromagnetic spectrum and the capabilities of detection equipment. As seen in the provided solution, the radiation of \( 10.0 \, Å \) is within the detectable range for X-rays, hence it is something an X-ray detector could identify.

Electromagnetic Wave Properties

Electromagnetic waves carry energy through space and possess a set of key properties: wavelength, frequency, velocity (speed of light in this case), amplitude, and polarization. The properties of these waves are interconnected; a change in one often results in a change in others. For example, increasing the energy of the wave usually decreases its wavelength due to the inversely proportional wavelength-frequency relationship.

Electromagnetic waves differ from other waves because they do not require a medium to travel through; they can move through the vacuum of space. This quality allows us to receive light from the sun and other astronomical objects. In the exercise, understanding the properties and the mathematics behind them enables the calculation of frequency from a known wavelength and vice versa.

From the perspective of problem-solving or practical applications, knowing these properties can help in the design of communications systems, medical imaging devices, and even in making sense of cosmic phenomena. By mastering the electromagnetic properties, students are better prepared to tackle real-world problems that involve these waves.