Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 40: Problem 54

Consider a 42.58 -MHz photon needed to produce NMR transition in free protons in a magnetic field of \(1.000 \mathrm{~T}\). What is the wavelength of the photon, its energy, and the region of the spectrum in which it lies? Could it be harmful to the human body?

Short Answer

Expert verified
Answer: The wavelength of the photon is 7.045 × 10^-3 m (7.045 mm), its energy is 2.823 × 10^-26 J, it belongs to the radio wave region of the electromagnetic spectrum, and it is not generally harmful to the human body.

Step by step solution

01

Find the wavelength of the photon

To find the wavelength of the photon, we will use the formula: speed of light (c) = frequency (f) × wavelength (λ) We have the frequency (f) = 42.58 MHz = 42.58 × 10^6 Hz, and the speed of light (c) = 3.00 × 10^8 m/s. Rearranging the formula for wavelength, we get: wavelength (λ) = c / f Now plug in the values: λ = (3.00 × 10^8 m/s) / (42.58 × 10^6 Hz) Calculating the wavelength: λ ≈ 7.045 × 10^-3 m The wavelength of the photon is 7.045 × 10^-3 m, or 7.045 mm.
02

Calculate the energy of the photon

To find the energy, we will use the formula: energy (E) = Planck's constant (h) × frequency (f) Planck's constant (h) approximately equals 6.63 × 10^-34 Js, and the frequency (f) is 42.58 × 10^6 Hz. Now plug in the values: E = (6.63 × 10^-34 Js) × (42.58 × 10^6 Hz) Calculating the energy: E ≈ 2.823 × 10^-26 J The energy of this photon is 2.823 × 10^-26 J.
03

Identify the region of the electromagnetic spectrum

The wavelength of the photon is 7.045 × 10^-3 m, or 7.045 mm. Comparing this wavelength to the electromagnetic spectrum, we can find the region it belongs to. Since the value for the wavelength is between 1 mm and 100 mm, it is located in the radio wave region of the electromagnetic spectrum.
04

Determine the potential harm to human body

The photon is in the radio wave region of the electromagnetic spectrum, which is a non-ionizing form of radiation. This type of radiation does not have enough energy to remove tightly bound electrons from atoms, and thus, generally does not have enough energy to damage DNA or cause harmful effects to humans. In typical NMR experiments, the exposure to radio waves is minimal, and the intensity of the radiation is low, so we consider this photon emission to be safe for the human body. In summary: - Wavelength: 7.045 × 10^-3 m (7.045 mm) - Energy: 2.823 × 10^-26 J - Region of the spectrum: Radio waves - Harmful to the human body: No

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

\(^{214} \mathrm{Pb}\) has a half-life of \(26.8 \mathrm{~min}\). How many minutes must elapse for \(90.0 \%\) of a given sample of \({ }^{214} \mathrm{~Pb}\) atoms to decay?

In neutron stars, which are roughly \(90 \%\) neutrons and supported almost entirely by nuclear forces, which of the following binding-energy terms becomes relatively dominant compared to ordinary nuclei? a) the Coulomb term b) the asymmetry term c) the pairing term d) all of the above e) none of the above

Consider the following fusion reaction, which allows stars to produce progressively heavier elements: \({ }_{2}^{3} \mathrm{He}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{4}^{7} \mathrm{Be}+\gamma\). The mass of \({ }_{2}^{3} \mathrm{He}\) is \(3.016029 \mathrm{u}\), the mass of \({ }_{2}^{4}\) He is \(4.002603 \mathrm{u}\), and the mass of \({ }_{4}^{7} \mathrm{Be}\) is \(7.0169298 \mathrm{u}\). The atomic mass unit is \(u=1.66 \cdot 10^{-27} \mathrm{~kg} .\) Assuming the Be atom is at rest after the reaction and neglecting any potential energy between the atoms and kinetic energy of the He nuclei, calculate the minimum possible energy and maximum possible wavelength of the photon \(\gamma\) that is released in this reaction.

The mean lifetime for a radioactive nucleus is \(4300 \mathrm{~s}\) What is its half-life?

Refer to the subsection "Terrestrial Fusion" in Section 40.4 to see how achieving controlled fusion would be the solution to mankind's energy problems, and how difficult it is to do. Why is it so hard? The Sun does it all the time (see the previous subsection, "Stellar Fusion"). Do we need to understand better how the Sun works to build a useful nuclear fusion reactor?
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Solid State Physics

Read Explanation

Dynamics

Read Explanation

Mechanics and Materials

Read Explanation

Energy Physics

Read Explanation

Wave Optics

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