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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.

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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.

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Most popular questions from this chapter

Take the size of a Rydberg atom to be the diameter of the orbit of the excited electron. If the researchers want to perform this experiment with the rubidium atoms in a gas, with atoms separated by a distance 10 times their size, the density of atoms per cubic centimeter should be about (a) \(10^{5}\) atoms \(/ \mathrm{cm}^{3} ;\) (b) \(10^{8}\) atoms \(/ \mathrm{cm}^{3} ;\) (c) \(10^{11}\) atoms \(/ \mathrm{cm}^{3}\) (d) \(10^{21}\) atoms \(/ \mathrm{cm}^{3}\).

How many different possible electron states are there in the \(n=100, l=2\) subshell? (a) \(2 ;\) (b) \(100 ;\) (c) 10,\(000 ;\) (d) 10 .

Consider the seventh excited level of the hydrogen atom. (a) What is the energy of this level? (b) What is the largest magnitude of the orbital angular momentum? (c) What is the largest angle between the orbital angular momentum and the z-axis?

The normalized radial wave function for the \(2 p\) state of the hydrogen atom is \(R_{2 p}=\left(1 / \sqrt{24 a^{5}}\right) r e^{-r / 2 a} .\) After we average over the angular variables, the radial probability function becomes \(P(r) d r=\left(R_{2 p}\right)^{2} r^{2} d r .\) At what value of \(r\) is \(P(r)\) for the \(2 p\) state a maximum? Compare your results to the radius of the \(n=2\) state in the Bohr model.

CP Classical Electron Spin. (a) If you treat an electron as a classical spherical object with a radius of \(1.0 \times 10^{-17} \mathrm{~m}\), what angular speed is necessary to produce a spin angular momentum of magnitude \(\sqrt{\frac{3}{4}} h ?\) (b) Use \(v=r \omega\) and the result of part (a) to calculate the speed \(v\) of a point at the electron's cquator. What does your result suggest about the validity of this model?

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