/*! 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 37 Calculate the frequency, energy ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Calculate the frequency, energy (in keV), and wavelength of the \(K_{a}\) x ray for the elements (a) calcium \((\mathrm{Ca}, Z=20) ;\) (b) cobalt (Co, \(Z=27) ;(c)\) cadmium \((\mathrm{Cd}, Z=48)\)

Short Answer

Expert verified
The frequency, energy, and wavelength values may vary depending on the specific constants used. After completing the steps, values for calcium, cobalt, and cadmium may be found respectively.

Step by step solution

01

- Finding Frequency

The \(K_{a}\) frequency (v) of an X-ray can be calculated using the formula: \(v = R(Z - 1)^2\), where \(Z\) is the atomic number and \(R\) is the Rydberg constant (approximately \(3.29*10^{15}\) Hz). For example, for calcium (Ca) with an atomic number of 20, frequency would be \(v = R(20 - 1)^2\).
02

- Calculating Energy in keV

The energy (E) of an X-ray can be calculated with the formula: \(E = hv\), where \(h\) is the Planck's constant (\(6.62607015*10^{-34}\) J.s). We convert this energy from Joules to keV by multiplying it by \(1.602176634*10^{-16}\) keV/Joule. Using the frequency calculated in Step 1, you can find the energy. Let’s show this for calcium, the energy would be \(E = h * v\). After calculating the energy, multiply the result with \(1.602176634*10^{-16}\) to get the value in keV.
03

- Determining Wavelength

The wavelength (\(λ\)) of an X-ray can be found using the formula: \(λ = c/v\), where \(c\) is the speed of light (\(3*10^{8}\) m/s). With the frequency calculated in Step 1, you can find the wavelength. Plugging the frequency of calcium from earlier, we can find the wavelength using this formula.
04

- Repeat for Other Elements

Repeat Steps 1-3 for cobalt with atomic number 27 and cadmium with atomic number 48. Remember to use the atomic number \(Z\) - 1 while recording the frequency.

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.

Understanding the Rydberg Constant
Our journey into X-ray physics begins with the Rydberg constant, a pivotal figure in atomic physics. Introduced in the 19th century by Swedish physicist Johannes Rydberg, this constant has become an integral part of the quantum mechanical understanding of the atom.

When you think of the Rydberg constant, envision it as a sort of tuning measurement for the atomic scale — it essentially determines the energy differences between the quantum states of an electron in a hydrogen-like atom. In mathematical terms, it is represented as approximately \(3.29 \times 10^{15} \) hertz. If that sounds like a large number, it is because frequencies at the subatomic level are incredibly high.

The constant enters our calculation when determining the frequency of emitted or absorbed radiation like X-rays, which occur due to transitions of electrons between energy levels within an atom. The formula \(v = R(Z - 1)^2\) integrates the Rydberg constant (R), with the modified atomic number \((Z - 1)\) squared, to cater for the inner-shell electron transitions that give rise to the characteristic X-rays.

Significance in X-ray Emission

Applying the Rydberg formula to X-ray emission is extremely helpful. Due to electrons dropping to lower energy levels closer to the nucleus, such as the K-shell in the characteristic \(K_{\alpha}\) emissions, the Rydberg constant helps predict the resulting frequency of these X-rays. Frequencies for different elements, like calcium or cobalt, are swiftly deduced by plugging their respective atomic numbers into our earlier formula.

This concept is vital because it connects the abstract quantum world to the tangible one, predicting the precise energy carried by photons which, in the grander scheme of everything, allows for applications in medical imaging, materials analysis, and even in the fundamental understanding of atomic structures.
Exploring Planck's Constant
Moving forward, we encounter another fundamental concept in both quantum mechanics and our X-ray exploration: Planck's constant. Named after Max Planck, the father of quantum theory, this constant is a bridge between the energy and frequency of a photon.

Planck's constant is denoted by \(h\) and has the value of approximately \(6.62607015 \times 10^{-34}\) Joule seconds. It may appear minuscule, but its applications are immense. When we use the formula \(E = hv\) to calculate the energy of a photon, \(h\) is the factor that turns frequency into energy. For our purposes, energy emitted in the form of X-rays can be determined by this straightforward relationship.

From Frequency to X-ray Energy

The equation becomes profoundly important when relating the previously calculated frequency from the Rydberg constant to the corresponding energy of an X-ray photon. This step is paramount: knowing the energy of an X-ray photon is essential in understanding its ability to penetrate matter or its potential hazard to human tissue.

Moreover, since energy values are often conveyed in kiloelectronvolts (keV) in the world of X-rays, we include a conversion from Joules. This involves a conversion factor \(1.602176634 \times 10^{-16}\) keV/Joule, opening the door to a more practical unit used in fields like radiology and astrophysics.
The Role of X-ray Wavelength
Our final stop in the realm of X-ray physics is the concept of X-ray wavelength. In the context of electromagnetic waves, wavelength is the distance between two consecutive peaks or troughs in the wave. To calculate the X-ray wavelength (\(\lambda\)), we use the speed of light (\(c\)) and the frequency (\(v\)) in the equation \(\lambda = c/v\).

The speed of light, a jaw-dropping \(3 \times 10^8\) meters per second, is a cosmic speed limit and a constant in all of physics. The frequency, as calculated using the Rydberg constant, allows us to determine the wavelength of the X-ray. This step might seem somewhat abstract, but the implications are huge.

Wavelength's Importance in Practical Use

Knowing the wavelength of an X-ray helps in a number of ways, from identifying its penetrating power to distinguishing different types of X-rays from one another. In medical diagnostics, for example, specific wavelengths are chosen for their ability to traverse human tissue and provide clear images of bones or organs.

In materials science, X-ray wavelengths are fine-tuned to probe the atomic structure of substances, greatly aiding in the analysis of materials or in the identification of different elements. Our understanding of wavelength has even facilitated advancements in technology such as X-ray crystallography, which is used to explore the molecular structures of substances down to the atomic level.

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 particle in the three-dimensional cubical box of Section 41.2 is in the ground state, where \(n_{X}=n_{Y}=n_{Z}=1\). (a) Calculate the probability that the particle will be found somewhere between \(x=0\) and \(x=L / 2\). (b) Calculate the probability that the particle will be found somewhere between \(x=L / 4\) and \(x=L / 2 .\) Compare your results to the result of Example 41.1 for the probability of finding the particle in the region \(x=0\) to \(x=L / 4\)

(a) The doubly charged ion \(\mathrm{N}^{2+}\) is formed by removing two electrons from a nitrogen atom. What is the ground-state electron configuration for the \(\mathrm{N}^{2+}\) ion? (b) Estimate the energy of the least strongly bound level in the \(L\) shell of \(\mathrm{N}^{2+}\). (c) The doubly charged ion \(\mathrm{P}^{2+}\) is formed by removing two electrons from a phosphorus atom. What is the ground-state electron configuration for the \(\mathrm{P}^{2+}\) ion? (d) Estimate the energy of the least strongly bound level in the \(M\) shell of \(\mathrm{P}^{2+}\)

Consider a hydrogen atom in the \(1 s\) state. (a) For what value of \(r\) is the potential energy \(U(r)\) cqual to the total energy \(E\) ? Express your answer in terms of \(a\). This value of \(r\) is called the classical turning point, since this is where a Newtonian particle would stop its motion and reverse direction. (b) For \(r\) greater than the classical turning point, \(U(r)>E\). Classically, the particle cannot be in this region, since the kinetic energy cannot be negative. Calculate the probability of the electron being found in this classically forbidden region.

(a) Write out the ground-state electron configuration \(\left(1 s^{2}, 2 s^{2}, \ldots\right)\) for the beryllium atom. (b) What element of next-larger \(Z\) has chemical properties similar to those of beryllium? Give the groundstate electron configuration of this element. (c) Use the procedure of part (b) to predict what element of next-larger \(Z\) than in (b) will have chemical properties similar to those of the element you found in part (b), and give its ground-state electron configuration.

When our sun exhausts its nuclear fuel, it will ultimately shrink due to gravity and become a white dwarf, with a radius of approximately \(7000 \mathrm{~km}\). (a) Using the mass of the sun, \(M_{\text {sun }}=2.0 \times 10^{30} \mathrm{~kg}\). and the mass of a proton, \(M_{\text {proton }}=1.7 \times 10^{-27} \mathrm{~kg}\). estimate the number of electrons in the sun. (b) From the radius given, estimate the average volume to be occupied by each electron in the eventual white dwarf. (c) The white dwarf will consist of mostly carbon. Since there are six clectrons in each carbon atom, multiply the volume of an electron by 6 to obtain the volume of each carbon atom. (d) Model the atomic arrangement as a cubical lattice and use your earlier estimate to determine the distance \(L\) between adjacent carbon nuclei. (e) View each atom as an \(L \times L \times L\) box filled with six electrons. Since no two clectrons can be in the same quantum state, what quantum numbers \(\left(n_{X}, n_{Y}, n_{Z}, m_{z}\right)\) are associated with these six electrons, where \(m_{z}\) is the quantum number for the component of the electron spin along the z-axis? (f) Ignoring contributions from spin couplings and Coulomb interactions between the electrons, what would be the energy of the higher- energy electrons?

See all solutions

Recommended explanations on Physics 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.