/*! 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 5 An excited nucleus emits a gamma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 38: Problem 5

An excited nucleus emits a gamma-ray photon with an energy of 2.45 \(\mathrm{MeV}\) . (a) What is the photon frequency?(b) What is the photon wavelength? (c) How does the wavelength compare with a typical nuclear diameter of \(10^{-14} \mathrm{m} ?\)

Short Answer

Expert verified
(a) Frequency is approximately \(5.92 \times 10^{20} \text{ Hz}\), (b) Wavelength is approximately \(5.07 \times 10^{-13} \text{ m}\), (c) Wavelength is about 50 times the nuclear diameter.

Step by step solution

01

Convert Energy to Frequency Formula

Use the formula that relates energy and frequency of a photon: \(E = h \cdot f\), where \(E\) is energy, \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J}\cdot\text{s}\), and \(f\) is frequency.
02

Compute Photon Frequency

Given the energy \(E = 2.45 \text{ MeV}\), first convert this energy into joules: \(2.45 \text{ MeV} = 2.45 \times 1.602 \times 10^{-13} \text{ J} = 3.926 \times 10^{-13} \text{ J}\). Then substitute this into the energy formula: \(f = \frac{E}{h} = \frac{3.926 \times 10^{-13} \text{ J}}{6.626 \times 10^{-34} \text{ J}\cdot\text{s}} \approx 5.92 \times 10^{20} \text{ Hz}\).
03

Use the Frequency-Wavelength Relationship

Use the formula \(c = \lambda \cdot f\), where \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\), to find the wavelength \(\lambda\). Rearrange it to \(\lambda = \frac{c}{f}\).
04

Calculate Photon Wavelength

Substitute the calculated frequency \(f = 5.92 \times 10^{20} \text{ Hz}\) into the formula: \(\lambda = \frac{3.00 \times 10^8 \text{ m/s}}{5.92 \times 10^{20} \text{ Hz}} \approx 5.07 \times 10^{-13} \text{ m}\).
05

Compare Wavelength to Nuclear Diameter

The wavelength found is \(\lambda = 5.07 \times 10^{-13} \text{ m}\), while a typical nuclear diameter is \(10^{-14} \text{ m}\). This means the photon wavelength is about 50 times longer than the nuclear diameter.

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 91影视!

Key Concepts

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

Gamma-ray Photon
Gamma-ray photons are highly energetic electromagnetic waves emitted by excited atomic nuclei during nuclear decay processes. These are among the most energetic forms of radiation and are often associated with nuclear physics and cosmic phenomena.

  • Energy Levels: Gamma-ray photons have energies typically greater than that of X-rays, and can be in the range of kiloelectronvolts (keV) to megaelectronvolts (MeV).
  • Nuclear Transitions: When a nucleus in an excited state transitions to a lower energy state, it releases energy in the form of a gamma-ray photon.
  • Applications: Gamma-rays are used in medical imaging, cancer treatments, and astrophysical observations.
Understanding gamma-ray photons is crucial for comprehending many phenomena in nuclear physics.
Photon Frequency
Photon frequency is a key concept in understanding electromagnetic radiation like light, X-rays, and gamma-rays. The frequency of a photon determines its color or type of electromagnetic radiation.

  • Definition: Frequency ( \(f\) ) refers to the number of cycles or oscillations of an electromagnetic wave per unit of time, usually measured in hertz (Hz).
  • Relation to Energy: Frequency is directly related to the energy of a photon by Planck鈥檚 equation: \(E = h \cdot f\) , where \(E\) is energy, \(h\) is Planck鈥檚 constant, and \(f\) is frequency.
  • Significance: A higher frequency implies a higher energy photon, making gamma-rays extremely high-frequency, high-energy photons.
Calculating the frequency of a photon helps in determining its energy and understanding its potential impact or applications.
Photon Wavelength
Photon wavelength is another fundamental property that describes electromagnetic waves. It is inversely proportional to the frequency, meaning as the frequency increases, the wavelength decreases.

  • Definition: Wavelength (\(\lambda\) ) is the distance between consecutive crests or troughs of a wave. It is typically measured in meters.
  • Frequency-Wavelength Relationship: Expressed by the equation \(c = \lambda \cdot f\), where \(c\) is the speed of light, showing that wavelength and frequency are inversely related.
  • Physical Implication: Increased energy (as in gamma-rays) results in decreased wavelength, making such wavelengths significantly smaller than typical nuclear dimensions.
Photon wavelength provides insight into the nature and capabilities of the electromagnetic wave.
Nuclear Physics
Nuclear physics is the branch of physics that studies the constituents and interactions of atomic nuclei. It has direct implications in understanding phenomena such as nuclear decay, which may involve gamma-ray emissions.

  • Core Concepts: Nuclear physics investigates the forces holding a nucleus together, energy transitions, and particle interactions, often involving the emission of gamma-rays.
  • Applications: This field is fundamental to nuclear energy generation, medical applications such as radiation therapy, and furthering our understanding of atomic synthesis in stars.
  • Relevance to Photons: Excited nuclei transitioning to lower energy states emit gamma-ray photons, linking nuclear physics to photon studies.
This field offers critical insights into the fundamental forces of nature and contributes to many practical and theoretical advancements.
Planck's Constant
Planck's constant (\(h\) ) is a fundamental constant used to describe the quantization of energy in quantum mechanics. It plays a crucial role in the relationship between a photon's energy and frequency.

  • Value: The constant is approximately \(6.626 \times 10^{-34} \text{ Js}\) (joule-seconds), giving a scale to quantum interactions.
  • Energy Relation: Planck鈥檚 constant is essential in the equation \(E = h \cdot f\), linking energy (\(E\) ) and frequency (\(f\) ) of a photon.
  • Significance: It is one of the cornerstones of quantum theory, influencing how we understand particle and wave duality.
Planck's constant helps bridge the classical and quantum worlds, allowing precise calculations of energy transformations in the quantum realm.

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

A \(4.78-\mathrm{MeV}\) alpha particle from a 226 \(\mathrm{Ra}\) decay makes a head-on collision with a uranium nucleus. A uranium nucleus has 92 protons. (a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uramium nucleus remains at rest and that the distance of closest approach is much greater than the radius of the uranium nucleus. (b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?

(a) A proton is moving at a speed much slower than the speed of light. It has kinetic energy \(K_{1}\) and momentum \(p_{1}\) . If the momentum of the proton is doubled, so \(p_{2}=2 p_{1},\) how is its new kinetic energy \(K_{2}\) related to \(K_{1} ?(b)\) A photon with energy \(E_{1}\) has momentum \(p_{1} .\) If another photon has momentum \(p_{2}\) that is twice \(p_{1},\) how is the energy \(E_{2}\) of the second photon related to \(E_{1} ?\)

A photon has momentum of magnitude \(8.24 \times 10^{-28} \mathrm{kg}\) . \(\mathrm{m} / \mathrm{s}\) (a) What is the energy of this photon? Give your answer in joules and in electron volts. (b) What is the wavelength of this photon? In what region of the electromagnetic spectrum does it lie?

\(X\) rays are produced in a tube operating at 18.0 \(\mathrm{kV}\) . After cmerging from the tube, \(x\) rays with the minimum wavelength produced strike a target and are Compton-scattered through an angle of \(45.0^{\circ} .\) (a) What is the original \(x\) -ray wavelength? (b) What is the wavelength of the scattered \(x\) rays? (c) What is the energy of the scattered \(x\) rays (in electron volts)?

Sirius B. The brightest star in the sky is Sirius, the Dog Star. It is actually a binary system of two stars, the smaller one (Sirius B) being a white dwarf. Spectral analysis of Sirius B indicates that its surface temperature is \(24,000 \mathrm{K}\) and that it radiates energy at a total rate of \(1.0 \times 10^{25} \mathrm{W}\) . Assume that it behaves like an ideal blackbody. (a) What is the total radiated intensity of Sirius \(\mathrm{B}\) (b) What is the peak-intensity wavelength? Is this wave-length visible to humans? (c) What is the radius of Sirius B? Express your answer in kilometers and as a fraction of our sun's radius. (d) Which star radiates more total energy per second, the hot Sirius or the (relatively) cool sun with a surface temperature of 5800 \(\mathrm{K}\) ? To find out, calculate the ratio of the total power radiated by our sun to the power radiated by Sirius B.
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Fundamentals of Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Radiation

Read Explanation

Atoms and Radioactivity

Read Explanation

Mechanics and Materials

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.