/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 (a) What is the frequency of rad... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) What is the frequency of radiation that has a wavelength of \(10 \mu \mathrm{m},\) about the size of a bacterium? (b) What is the wavelength of radiation that has a frequency of \(5.50 \times 10^{14} \mathrm{~s}^{-1}\) ? (c) Would the radiations in part (a) or part (b) be visible to the human eye? (d) What distance does electromagnetic radiation travel in \(50.0 \mu \mathrm{s} ?\)

Short Answer

Expert verified
(a) The frequency of radiation with a wavelength of \(10\mu m\) is \(3\times10^{13} s^{-1}\). (b) The wavelength of radiation with a frequency of \(5.50\times10^{14} s^{-1}\) is \(5.45\times10^{-7}m\) or 545nm. (c) The radiation in part (a) is not visible to the human eye, while the radiation in part (b) is visible. (d) Electromagnetic radiation travels 15000 meters in 50.0μs.

Step by step solution

01

Part (a): Find the frequency of radiation with a wavelength of 10μm

Given the wavelength, λ = 10 μm = \(1\times10^{-5}\) m, and the speed of light, c = \(3\times10^8\) m/s, we can find the frequency (v) using the formula: Frequency (v) = Speed of light (c) / Wavelength (λ) v = \(\frac{3\times10^8}{1\times10^{-5}}\) v = \(3\times10^{13}\) s\(^{-1}\) So, the frequency of radiation with a wavelength of 10μm is \(3\times10^{13}\) s\(^{-1}\).
02

Part (b): Find the wavelength of radiation with a frequency of \(5.50\times10^{14}\) s\(^{-1}\)

Given the frequency, v = \(5.50\times10^{14}\) s\(^{-1}\), and the speed of light, c = \(3\times10^8\) m/s, we can find the wavelength (λ) using the formula: Wavelength (λ) = Speed of light (c) / Frequency (v) λ = \(\frac{3\times10^8}{5.50\times10^{14}}\) λ = \(5.45\times10^{-7}\) m So, the wavelength of radiation with a frequency of \(5.50\times10^{14}\) s\(^{-1}\) is \(5.45\times10^{-7}\) m or 545nm.
03

Part (c): Determine if the radiations are visible to the human eye

The visible range of wavelengths for the human eye is approximately 400nm to 700nm. For part (a), the wavelength is 10μm, which is equal to 10,000nm. This is not within the visible range. For part (b), the wavelength is 545nm, which is within the visible range. Therefore, the radiation in part (a) is not visible to the human eye, while the radiation in part (b) is visible.
04

Part (d): Find the distance electromagnetic radiation travels in 50.0μs

To find the distance electromagnetic radiation travels, we can use the formula: distance (d) = speed of light (c) × time (t). Given the time, t = 50.0μs = \(5\times10^{-5}\) s, and the speed of light, c = \(3\times10^8\) m/s: distance (d) = \((3\times10^8) \times (5\times10^{-5})\) distance (d) = 15000 m So, electromagnetic radiation travels 15000 meters in 50.0μs.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wavelength and Frequency Relationship
The relationship between wavelength (represented by the Greek letter lambda, \(\lambda\)) and frequency (represented by the Greek letter nu, \(u\)) is one of the fundamental principles in understanding electromagnetic radiation. These two properties are inversely proportional, meaning that as the wavelength increases, the frequency decreases, and vice versa. This is expressed by the formula \(u = \frac{c}{\lambda}\) where \(c\) is the speed of light in a vacuum, which is approximately \(3 \times 10^8 \, \text{m/s}\).

When students tackle problems involving the calculation of one variable given the other, understanding this inverse relationship is key. For example, if a radiation has a longer wavelength, it would inherently have a lower frequency, and this is crucial when classifying different types of electromagnetic radiation on the spectrum.
Visible Light Spectrum
The visible light spectrum represents a small portion of the electromagnetic spectrum that can be detected by the human eye. This range typically extends from about 400 nanometers (nm) to 700 nm. Within this range, different wavelengths correspond to different colors, with violet on the shorter wavelength end (approximately 400 nm) and red on the longer wavelength end (approximately 700 nm).

For educational purposes, it's important to illustrate that any electromagnetic radiation with a wavelength outside of this range is invisible to us. As shown in the exercise solutions, a wavelength of \(10 \mu\text{m}\), or 10,000 nm, is far beyond the scope of human vision, while a wavelength of 545 nm falls within the visible spectrum, corresponding to the color green.
Speed of Light
A constant that often appears in physics formulas is the speed of light, denoted by \(c\). This intrinsic factor of the universe is approximately \(3 \times 10^8 \, \text{m/s}\) in a vacuum. This constant is not just a speed limit for light but is also the speed at which all electromagnetic waves propagate in a vacuum.

Students should realize that the speed of light remains the same regardless of the observer's perspective or the source of light. This principle holds true for all electromagnetic radiation calculations, influencing how we perceive time and distance in the universe. It's the central element in calculating the distance electromagnetic waves travel over a given time, just like in the exercise where radiation travels 15,000 meters in 50.0 microseconds.
Electromagnetic Spectrum
The electromagnetic (EM) spectrum encompasses all types of electromagnetic radiation, which vary in wavelength and frequency. At one end of the spectrum, we have gamma rays with very short wavelengths and high frequencies, and at the other end, we have radio waves with long wavelengths and low frequencies. Between these extremes lie other types of EM radiation including X-rays, ultraviolet (UV) light, visible light, infrared (IR) light, and microwaves.

Understanding the EM spectrum is crucial for grasping how different types of radiation interact with matter and its applications across various technologies. For instance, microwaves heat our food, X-rays help in medical diagnostics, and infrared technology is used in night-vision devices. The textbook exercise reinforces the concept that not all radiation is visible and reminds us of the vast range of EM waves surrounding us.

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

Certain elements emit light of a specific wavelength when they are burned. Historically, chemists used such emission wavelengths to determine whether specific elements were present in a sample. Characteristic wavelengths for some of the elements are given in the following table: \(\begin{array}{llll}\mathrm{Ag} & 328.1 \mathrm{nm} & \mathrm{Fe} & 372.0 \mathrm{nm} \\ \mathrm{Au} & 267.6 \mathrm{nm} & \mathrm{K} & 404.7 \mathrm{nm} \\ \mathrm{Ba} & 455.4 \mathrm{nm} & \mathrm{Mg} & 285.2 \mathrm{nm} \\ \mathrm{Ca} & 422.7 \mathrm{nm} & \mathrm{Na} & 589.6 \mathrm{nm} \\ \mathrm{Cu} & 324.8 \mathrm{nm} & \mathrm{Ni} & 341.5 \mathrm{nm}\end{array}\) (a) Determine which elements emit radiation in the visible part of the spectrum. (b) Which element emits photons of highest energy? Of lowest energy? (c) When burned, a sample of an unknown substance is found to emit light of frequency \(6.59 \times 10^{14} \mathrm{~s}^{-1} .\) Which of these elements is probably in the sample?

Indicate whether energy is emitted or absorbed when the following electronic transitions occur in hydrogen: (a) from \(n=2\) to \(n=6,\) (b) from an orbit of radius \(4.76 \AA\) to one of radius \(0.529 \AA,(\mathrm{c})\) from the \(n=6\) to the \(n=9\) state.

(a) Calculate the energies of an electron in the hydrogen atom for \(n=1\) and for \(n=\infty .\) How much energy does it require to move the electron out of the atom completely (from \(n=1\) to \(n=\infty),\) according to Bohr? Put your answer in \(\mathrm{kJ} / \mathrm{mol}\). (b) The energy for the process \(\mathrm{H}+\) energy \(\rightarrow \mathrm{H}^{+}+\mathrm{e}^{-}\) is called the ionization energy of hydrogen. The experimentally determined value for the ionization energy of hydrogen is \(1310 \mathrm{~kJ} / \mathrm{mol}\). How does this compare to your calculation?

Bohr's model can be used for hydrogen-like ions -ions that have only one electron, such as \(\mathrm{He}^{+}\) and \(\mathrm{Li}^{2+}\). (a) Why is the Bohr model applicable to \(\mathrm{He}^{+}\) ions but not to neutral He atoms? (b) The ground-state energies of \(\mathrm{H}, \mathrm{He}^{+},\) and \(\mathrm{Li}^{2+}\) are tabulated as follows: By examining these numbers, propose a relationship between the ground-state energy of hydrogen-like systems and the nuclear charge, \(Z\). (c) Use the relationship you derive in part (b) to predict the ground-state energy of the \(\mathrm{C}^{5+}\) ion.

If you put 120 volts of electricity through a pickle, the pickle will smoke and start glowing orange-yellow. The light is emitted because sodium ions in the pickle become excited; their return to the ground state results in light emission. (a) The wavelength of this emitted light is \(589 \mathrm{nm} .\) Calculate its frequency. (b) What is the energy of 0.10 mole of these photons? (c) Calculate the energy gap between the excited and ground states for the sodium ion. (d) If you soaked the pickle for a long time in a different salt solution, such as strontium chloride, would you still observe \(589-\mathrm{nm}\) light emission? Why or why not?

See all solutions

Recommended explanations on Chemistry Textbooks

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.