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 ps?

Short answer

The frequency of radiation with a wavelength of $0.86\mathrm{nm}$ is approximately $3.49\times {10}^{17}$ s${}^{-1}$. The wavelength of radiation with a frequency of $6.4\times {10}^{11}$ s${}^{-1}$ is approximately $4.69\times {10}^5$ nm. The radiation from part (a) would be detected by an X-ray detector, while the radiation from part (b) would not be detected. Electromagnetic radiation travels a distance of approximately $1.14\times {10}^{-4}$ m in 0.38 ps.

Step-by-step solution

Write down the given information

We are given the wavelength of a photon as 0.86 nm (nanometers).

Convert the wavelength to meters

In order to use the formula $c=\lambda \nu$, we need to convert the wavelength from nanometers to meters, using the conversion factor (1 nm = ${10}^{-9}$ m). So, $\lambda =0.86\times {10}^{-9}$ m.

Calculate the frequency

Using the formula $c=\lambda \nu$, we can solve for the frequency ($\nu$) by dividing both sides by the wavelength ($\lambda$): $\nu =\frac{c}{\lambda }$ Now, substitute the values of the speed of light ($c$) and the wavelength ($\lambda$) in meters: $\nu =\frac{3\times {10}^8}{0.86\times {10}^{-9}}$

Calculate the result

After performing the calculations, we find the frequency: $\nu \approx 3.49\times {10}^{17}$ s${}^{-1}$. #b. Calculating the wavelength of radiation given the frequency#

Write down the given information

We are given the frequency of a photon as $6.4\times {10}^{11}$ s${}^{-1}$ (inverse seconds, or Hz).

Calculate the wavelength

Using the formula $c=\lambda \nu$, we can solve for the wavelength ($\lambda$) by dividing both sides by the frequency ($\nu$): $\lambda =\frac{c}{\nu }$ Now, substitute the values of the speed of light ($c$) and the given frequency ($\nu$): $\lambda =\frac{3\times {10}^8}{6.4\times {10}^{11}}$

Calculate the result

After performing the calculations, we find the wavelength in meters: $\lambda \approx 4.69\times {10}^{-4}$ m.

Convert the result back to nanometers

Convert the wavelength back to nanometers using the conversion factor (1 m = ${10}^9$ nm): $\lambda =4.69\times {10}^{-4}\times {10}^9$ nm $\lambda \approx 4.69\times {10}^5$ nm #c. Determining if the radiation will be detected by an X-ray detector#

Compare the wavelengths to the range detected by an X-ray detector

We know that X-ray detectors can detect wavelengths in the range of 0.1 nm to 10 nm. - In part (a), the wavelength was 0.86 nm, which falls within the range of an X-ray detector. Thus, this radiation would be detected. - In part (b), the wavelength was approximately $4.69\times {10}^5$ nm, which is outside the range of an X-ray detector. Thus, this radiation would not be detected. #d. Distance traveled by electromagnetic radiation in 0.38 ps#

Write down the given information

We are given the time traveled by electromagnetic radiation as 0.38 ps (picoseconds).

Convert the time to seconds

We need to convert the time from picoseconds to seconds, using the conversion factor (1 ps = ${10}^{-12}$ s). Therefore, $t=0.38\times {10}^{-12}$ s.

Calculate the distance traveled

To calculate the distance traveled, we use the formula $d=ct$, where $d$ is the distance, $c$ is the speed of light, and $t$ is the time. Now, substitute the values of the speed of light ($c$) and the time ($t$) in seconds: $d=(3\times {10}^8)(0.38\times {10}^{-12})$

Calculate the result

After performing the calculations, we find the distance traveled: $d\approx 1.14\times {10}^{-4}$ m.

Key concepts

Frequency Calculation

Frequency is a fundamental concept in understanding electromagnetic radiation. It refers to the number of times a wave oscillates or passes through a point in one second. This is measured in hertz (Hz), which is equivalent to s${}^{-1}$. To calculate the frequency of electromagnetic radiation, we use the equation $c=\lambda u$, where $c$ is the speed of light, $\lambda$ is the wavelength, and $u$ is the frequency.

In our exercise, we were given the wavelength as 0.86 nm (nanometers). To perform any calculations, it's important to convert this into meters, since the speed of light is commonly expressed in meters per second. By converting 0.86 nm into meters, we determine $\lambda =0.86\times {10}^{-9}$ m. Using the rearranged formula $u=\frac{c}{\lambda }$, we substitute the known values. This allows us to find the frequency, which is approximately $3.49\times {10}^{17}$ Hz.

Understanding frequency is crucial for analyzing different forms of electromagnetic radiation, as it gives us insight into the energy and behavior of the waves.

Wavelength Calculation

Wavelength calculation is essential for understanding the properties of electromagnetic waves. Wavelength, typically measured in meters, is the distance between two consecutive peaks of a wave. Similar to frequency, it is a key factor that determines the behavior and energy of the radiation.

The equation used to calculate wavelength from frequency is $c=\lambda u$. Solving for the wavelength gives us $\lambda =\frac{c}{u}$. In the exercise, we were given a frequency of $6.4\times {10}^{11}$ s${}^{-1}$. By substituting into the formula, we find $\lambda \approx 4.69\times {10}^{-4}$ m.

To make this more relatable, we convert the wavelength back into nanometers, because nanometers are a more convenient scale for small distances like those found in electromagnetic waves. This gives us approximately $4.69\times {10}^5$ nm.

Knowing how to calculate wavelength is crucial for understanding and predicting how electromagnetic radiation will interact with materials and detectors.

X-ray Detector

X-ray detectors are specialized devices designed to capture and measure X-rays, a specific type of electromagnetic radiation. X-ray wavelengths typically range between 0.1 nm and 10 nm. These detectors are highly effective in capturing radiation within this range for applications like medical imaging or material analysis.

In our problem, we needed to determine whether the radiations from parts (a) and (b) would be captured by an X-ray detector.

• In part (a), the radiation had a wavelength of 0.86 nm, which falls within the X-ray detection range. Therefore, this radiation would be detected by an X-ray detector.
• In part (b), the wavelength was about $4.69\times {10}^5$ nm. This wavelength is far outside the range of what an X-ray detector can capture. Hence, it would not be detected.

Understanding the sensitivity range of detectors is important for correctly interpreting data from experiments involving electromagnetic radiation.

Speed of Light

The speed of light is a fundamental constant in physics, crucial for understanding the behavior of electromagnetic waves. Denoted by $c$, its value is approximately $3.00\times {10}^8$ m/s. This is the speed at which all electromagnetic radiation, including visible light, radio waves, and X-rays, travels through a vacuum.

The speed of light plays a central role in calculations involving electromagnetic waves, particularly when determining the relationship between frequency and wavelength. In the exercise, it was used to both calculate frequency from wavelength and vice versa.

Additionally, when calculating the distance electromagnetic radiation travels in a given time, we also use the speed of light with the formula $d=ct$. For example, if radiation travels for 0.38 ps, by converting this to seconds and multiplying by the speed of light, we find the radiation travels a distance of approximately $1.14\times {10}^{-4}$ m.

Having a good grasp of the speed of light's role allows for precise calculations and understanding of various electromagnetic wave phenomena.