Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 20

(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

Expert verified
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

01

Write down the given information

We are given the wavelength of a photon as 0.86 nm (nanometers).
02

Convert the wavelength to meters

In order to use the formula \(c = λν\), we need to convert the wavelength from nanometers to meters, using the conversion factor (1 nm = \(10^{-9}\) m). So, \(λ = 0.86 \times 10^{-9}\) m.
03

Calculate the frequency

Using the formula \(c = λν\), we can solve for the frequency (\(ν\)) by dividing both sides by the wavelength (\(λ\)): \(ν = \frac{c}{λ}\) Now, substitute the values of the speed of light (\(c\)) and the wavelength (\(λ\)) in meters: \(ν = \frac{3 \times 10^8}{0.86 \times 10^{-9}}\)
04

Calculate the result

After performing the calculations, we find the frequency: \(ν \approx 3.49 \times 10^{17}\) s\(^{-1}\). #b. Calculating the wavelength of radiation given the frequency#
05

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).
06

Calculate the wavelength

Using the formula \(c = λν\), we can solve for the wavelength (\(λ\)) by dividing both sides by the frequency (\(ν\)): \(λ = \frac{c}{ν}\) Now, substitute the values of the speed of light (\(c\)) and the given frequency (\(ν\)): \(λ = \frac{3 \times 10^8}{6.4 \times 10^{11}}\)
07

Calculate the result

After performing the calculations, we find the wavelength in meters: \(λ \approx 4.69 \times 10^{-4}\) m.
08

Convert the result back to nanometers

Convert the wavelength back to nanometers using the conversion factor (1 m = \(10^9\) nm): \(λ = 4.69 \times 10^{-4} \times 10^9\) nm \(λ \approx 4.69 \times 10^5\) nm #c. Determining if the radiation will be detected by an X-ray detector#
09

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#
10

Write down the given information

We are given the time traveled by electromagnetic radiation as 0.38 ps (picoseconds).
11

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.
12

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: \(d = (3 \times 10^8)(0.38 \times 10^{-12})\)
13

Calculate the result

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of \(325 \mathrm{nm}\) (a) What is the energy of a photon of this wavelength? (b) What is the energy of a mole of these photons? (c) How many photons are in a \(1.00 \mathrm{~mJ}\) burst of this radiation? \((\mathbf{d})\) These \(\mathrm{UV}\) photons can break chemical bonds in your skin to cause sunburn-a form of radiation damage. If the \(325-\mathrm{nm}\) radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in \(\mathrm{kJ} / \mathrm{mol}\).

Sketch the shape and orientation of the following types of orbitals: \((\mathbf{a}) s,(\mathbf{b}) p_{z},(\mathbf{c}) d_{x y}\)

(a) A green laser pointer emits light with a wavelength of \(532 \mathrm{nm}\). What is the frequency of this light? (b) What is the energy of one of these photons? (c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of \(532-\mathrm{nm}\) photons. What is the energy gap between the ground state and excited state in the laser material?

Consider a transition of the electron in the hydrogen atom from \(n=8\) to \(n=3\). (a) Is \(\Delta E\) for this process positive or negative? (b) Determine the wavelength of light that is associated with this transition. Will the light be absorbed or emitted? (c) In which portion of the electromagnetic spectrum is the light in part (b)?

Among the elementary subatomic particles of physics is the muon, which decays within a few microseconds after formation. The muon has a rest mass 206.8 times that of an electron. Calculate the de Broglie wavelength associated with a muon traveling at \(8.85 \times 10^{5} \mathrm{~cm} / \mathrm{s}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free