Question

(a) What is the frequency of radiation whose wavelength is \(0.86 \mathrm{~nm}\) ? (b) What is the wavelength of radiation that has a frequency of \(6.4 \times 10^{11} \mathrm{~s}^{-1}\) ? (c) Would the radiations in part (a) or part (b) be detected by an X-ray detector? (d) What distance does electromagnetic radiation travel in \(0.38 \mathrm{ps}\) ?

Short answer

(a) The frequency of radiation with a wavelength of 0.86 nm is approximately \(3.49 * 10^{17} \, s^{-1}\). (b) The wavelength of radiation with a frequency of \(6.4 * 10^{11} \, s^{-1}\) is approximately \(4.69 * 10^{-4}\, m\). (c) The radiation in part (a) would be detected by an X-ray detector, but the radiation in part (b) would not. (d) The distance traveled by electromagnetic radiation in 0.38 ps is approximately \(1.14 * 10^{-4}\, m\).

Step-by-step solution

Part (a) - Calculate the frequency of radiation given its wavelength

Given the wavelength of radiation (λ) = 0.86 nm, we need to find the frequency (f). First, we need to convert the wavelength to meters: \( \lambda = 0.86 nm * 10^{-9} m/nm = 0.86 * 10^{-9}m \) Next, use the equation relating the speed of light (c), wavelength (λ), and frequency (f): \( c = \lambda * f \) Now, we'll solve for the frequency (f): \( f = \frac{c}{\lambda} \) The speed of light (c) is approximately 3.0 x 10^8 m/s. Plug in the values of c and λ: \( f = \frac{3.0 * 10^8 \, m/s}{0.86 * 10^{-9}m} \)

Part (a) - Final answer

Calculating the frequency (f): \( f \approx 3.49 * 10^{17} \, s^{-1} \) So the frequency of the radiation with a wavelength of 0.86 nm is approximately 3.49 × 10^17 s^(-1).

Part (b) - Calculate the wavelength of radiation given its frequency

Given the frequency of radiation (f) = 6.4 × 10^11 s^(-1), we need to find the wavelength (λ). Using the equation relating the speed of light (c), wavelength (λ), and frequency (f): \( \lambda = \frac{c}{f} \) Plug in the values of c and f: \( \lambda = \frac{3.0 * 10^8 \, m/s}{6.4 * 10^{11} \, s^{-1}} \)

Part (b) - Final answer

Calculating the wavelength (λ): \( \lambda \approx 4.69 * 10^{-4} \, m \) So the wavelength of the radiation with a frequency of 6.4 × 10^11 s^(-1) is approximately \(4.69 * 10^{-4}\, m\).

Part (c) - Determine whether the radiations in part (a) or (b) would be detected by an X-ray detector

X-ray detectors typically detect radiation with wavelengths between 0.01 nm and 10 nm. Now, we need to compare the wavelengths of the radiations from part (a) and (b). In part (a), the wavelength is 0.86 nm, which is within the X-ray detector's detection range. In part (b), the wavelength is approximately \(4.69 * 10^{-4}\, m\), which is equal to 469 nm. This is outside the detection range of an X-ray detector. Thus, the radiation in part (a) would be detected by an X-ray detector, while the radiation in part (b) would not.

Part (d) - Calculate the distance traveled by electromagnetic radiation in a given time

Given the time (t) that the electromagnetic radiation travels = 0.38 ps, we need to find the distance traveled (d). First, we need to convert the time from picoseconds (ps) to seconds (s): \( t = 0.38 \, ps * 10^{-12} \, s/ps = 0.38 * 10^{-12} \, s \) Next, use the equation for distance (d) traveled by electromagnetic radiation in terms of speed of light (c) and time (t): \( d = c * t \) Plug in the values of c and t: \( d = 3.0 * 10^8 \, m/s * 0.38 * 10^{-12} \, s \)

Part (d) - Final answer

Calculating the distance (d): \( d \approx 1.14 * 10^{-4} \, m \) So the distance traveled by electromagnetic radiation in 0.38 ps is approximately \(1.14 * 10^{-4}\, m\).

Key concepts

Frequency Calculation

Frequency is a key concept in understanding electromagnetic radiation. It refers to the number of wave cycles that pass a point per unit time, usually measured in Hertz (Hz). To calculate the frequency of radiation when given its wavelength, we use the fundamental relationship between frequency (\( f \)), wavelength (\( \lambda \)), and the speed of light (\( c \)). The formula is:
\[ f = \frac{c}{\lambda} \]

• The speed of light \( c \) is a constant value, approximately \(3.0 \times 10^8 \) meters per second (m/s).
• Wavelength \( \lambda \) must be converted into meters for this equation to work correctly.

In our specific example, we began with a wavelength of \(0.86\, \text{nm}\) (nanometers), which converts to \(0.86 \times 10^{-9}\, \text{m}\).
By plugging these values into the formula, we find that the frequency of the radiation is approximately \(3.49 \times 10^{17}\, \text{s}^{-1}\), showing how often these waves occur every second.

Wavelength Conversion

Wavelength conversion is crucial when working with electromagnetic waves, especially when switching between different units or working with frequencies. To convert frequency back to wavelength, we use the equation:
\[ \lambda = \frac{c}{f} \]

• The formula involves the same constant speed of light \( c = 3.0 \times 10^8 \text{ m/s} \) and the given frequency \( f \).
• The frequency must be in seconds to ensure compatibility with the speed of light's units.

Given a frequency of \(6.4 \times 10^{11} \text{ s}^{-1}\), the calculation returns a wavelength of approximately \(4.69 \times 10^{-4}\, \text{m}\), or \(469\, \text{nm}\). This demonstrates converting between frequency and wavelength and illustrates how they inversely relate: as frequency increases, wavelength decreases.

X-ray Detection

X-ray detection involves the identification of electromagnetic radiation in the X-ray part of the spectrum. X-rays typically have wavelengths in the range of \(0.01\, \text{nm}\) to \(10\, \text{nm}\), which categorizes them as high-energy waves. The versatility of X-ray detection tools stems from their ability to perceive small objects or changes.
In our earlier example, the wavelength of radiation in part (a) was \(0.86\, \text{nm}\), positioning it well within the range of wavelengths detectable by X-ray detectors. Conversely, the wavelength in part (b), which was approximately \(4.69\, \times 10^{-4} \text{m}\) or \(469\, \text{nm},\) falls outside the uniquely short wavelength range characteristic of X-rays, making it undetectable by standard X-ray detectors.

Speed of Light

The speed of light is a fundamental constant central to all calculations relating to electromagnetic radiation. It is denoted as \( c \) and its value is approximately \(3.0 \times 10^8 \text{ m/s} \). This speed is crucial in determining both frequency and wavelength relationships:

• Connecting mechanical wave properties like speed, wavelength, and frequency
• Helping compute distances electromagnetic waves can traverse over time

For example, in part (d) of our exercise, we calculated how far electromagnetic radiation travels in a given time period, using:
\[ d = c \times t \]With a time period of \(0.38\, \text{ps}\) (picoseconds) or \(0.38 \times 10^{-12} \text{s}\), the resulting distance that light travels is approximately \(1.14 \times 10^{-4}\, \text{m}\). This result emphasizes the immense speed at which electromagnetic radiation, such as light, can cover vast distances in very short times.