Problem 20 - Chemistry The Central Science In Si Units

Question

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

Step-by-Step Solution

Answer

(a) 3.49 x 10^17 s^-1, (b) 469 µm, (c) Part (a) can be detected, (d) 114 µm.

Step 1: Understanding Radiation Frequency and Wavelength Relationship

To find the frequency of radiation, we can use the relationship between frequency \( \,f\) and wavelength \( \,\lambda\): \( c = \lambda f\), where \( c\) is the speed of light (approximately \( 3.00 \times 10^8 \, m/s\)).

Step 2: Calculating Frequency for Part (a)

Given that the wavelength \( \lambda = 0.86 \, nm = 0.86 \times 10^{-9} \, m\). The formula is rearranged to \( f = \frac{c}{\lambda}\): \[ f = \frac{3.00 \times 10^8}{0.86 \times 10^{-9}} \] Solve this to find \( f = 3.49 \times 10^{17} \, s^{-1}\).

Step 3: Calculating Wavelength for Part (b)

For the radiation with a frequency of \( 6.4 \times 10^{11} \, s^{-1}\), use the formula \( \lambda = \frac{c}{f}\): \[ \lambda = \frac{3.00 \times 10^8}{6.4 \times 10^{11}} \] Solving gives \( \lambda = 4.69 \times 10^{-4} \, m = 469 \, \mu m\).

Step 4: Determining Detectability with an X-ray Detector for Part (c)

X-ray detectors typically detect wavelengths roughly in the range of 0.01 nm to 10 nm. For part (a), the wavelength \( 0.86 \, nm\) falls within this range and can be detected.For part (b), the wavelength \( 469 \, \mu m = 469 \times 10^{-6} \, m = 469,000 \, nm\) is outside the range and cannot be detected.

Step 5: Calculating Distance Traveled for Part (d)

Given \( 0.38 \, ps\) which is \( 0.38 \times 10^{-12} \, s\), use \( \text{distance} = c \times \text{time}\):\[ \text{distance} = 3.00 \times 10^8 \, m/s \times 0.38 \times 10^{-12} \, s \] Calculate the distance to get \( 1.14 \times 10^{-4} \, m = 114 \mu m\).

Key Concepts

Frequency and Wavelength Relationship

In the world of electromagnetic radiation, frequency and wavelength share a unique relationship. These two properties help define the behavior of waves within the electromagnetic spectrum.
The relationship is expressed with the equation \( c = \lambda f \), where \( c \) stands for the speed of light, \( \lambda \) is the wavelength, and \( f \) is the frequency. This equation encapsulates that as one increases, the other decreases, indicating an inverse relationship.
To solve problems like the one given, you rearrange the formula depending on which property you need to find. For instance, finding frequency when wavelength is known involves altering the equation to \( f = \frac{c}{\lambda} \). Similarly, if you know the frequency, you can determine the wavelength via \( \lambda = \frac{c}{f} \).

• This equation is vital in fields like physics and engineering, making it essential for understanding everything from radio waves to X-rays.

Speed of Light

The speed of light is a constant in a vacuum, symbolized by \( c \). It is approximately \( 3.00 \times 10^8 \) meters per second.
This speed is pivotal because it links frequency and wavelength in the electromagnetic spectrum. All electromagnetic waves travel at this speed in a vacuum.
Light speed is not only fascinating because of its constancy, but it also represents the fastest speed at which information or matter can travel.

• When dealing with electromagnetic waves, always remember that \( c \) ties wavelength and frequency together consistently.
• This speed governs the propagation of light and helps in defining the principles of relativity.

X-ray Detection

X-ray detection involves capturing and interpreting electromagnetic radiation in the X-ray spectrum.
X-rays lie on the spectrum at wavelengths from about 0.01 nm to 10 nm. An X-ray detector specializes in identifying these waves, which are shorter and more energetic than visible light.
In the provided exercise, we see that the wavelength of 0.86 nm is suitable for detection as it lies within this range.

• X-rays are famous for medical imaging because of their ability to pass through soft tissues and offer a view of bones and other dense structures.
• In technology and security, X-rays are vital, making their detectability a crucial aspect of numerous applications.

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, ranging from extremely short wavelengths of gamma rays to long wavelengths of radio waves.
Understanding the spectrum is essential. Each type of wave, including radio waves, microwaves, visible light, and X-rays, plays distinct roles in technology and science.
The exercise reveals this spectrum's diversity by addressing X-rays and radio waves within its broad range.

• X-rays, as discussed, are high-energy waves crucial in medical imaging.
• Other parts of the spectrum, like radio waves, are vital for communications, showcasing the vast range of purposes these waves serve.

Encompassing all these waves highlights the dynamic nature and utility of the electromagnetic spectrum in our daily lives.