/*! 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 58 Using a beam of electrons accele... [FREE SOLUTION] | 91Ó°ÊÓ

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Using a beam of electrons accelerated in an X-ray tube, we wish to knock an electron out of the \(K\) shell of a given element in a target. Section 7.8 gives the energies in a hydrogenlike atom as \(Z^{2}\left(-13.6 \mathrm{eV} / \mathrm{n}^{2}\right)\). Assume that for \(f\) airly high \(Z\), a \(K\) -shell electron can be treated as orbiting the nucleus alone, (a) A typical accelerating potential in an X-ray tube is \(50 \mathrm{kV}\). In roughly how high a \(Z\) could a hole in the \(K\) -shell be produced? (b) Could a hole be produced in elements of higher \(Z\) ?

Short Answer

Expert verified
For a 50 kV accelerating voltage, atoms with an atomic number of approximately Z can have a hole produced in the K-shell.

Step by step solution

01

Convert accelerating potential to energy

The energy that an electron possesses is given by the product of its charge and the accelerating voltage. Knowing that we have a voltage of 50 kV and 1 eV = 1.6 x 10^-19 joules, the energy in eV can be found by \( E = 50 \times 10^3 eV = 50 \times 10^3 \times 1.6 \times 10^{-19} joules \)
02

Equate the potential energy to the energy formula of a hydrogenlike atom

Now equate this computed electron energy with the energy formula of a hydrogenlike atom, given as \(Z^{2}\left(-13.6 \mathrm{eV} / \mathrm{n}^{2}\right)\), where n is the shell number and should be 1 because we are dealing with K-shell electrons. So the equation becomes: \( 50 \times 10^3 = Z^2 \times \frac{-13.6}{1}\)
03

Solve for Z

Solving the equation for Z, we find the atomic number Z that could produce a hole in the K-shell. The answer should be a positive integer.

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Key Concepts

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

Electron Acceleration
When electrons in an X-ray tube are accelerated, they gain kinetic energy due to the electric potential they travel through. This process is crucial in X-ray tube physics, as it determines the energy and capabilities of the electrons to knock out inner-shell electrons from atoms.

The maximum energy an electron can gain while being accelerated in an X-ray tube is directly related to the accelerating voltage. If you have an accelerating potential of 50 kV, for example, the energy provided to each electron will be 50,000 electron volts (eV). This energy is calculated using the formula:
  • Energy = Charge of Electron × Accelerating Voltage
  • For an electron, this translates to: Energy = 1 e × 50,000 V
Given that 1 e is approximately equal to the charge of an electron (1.6 × 10^-19 C), the energy in joules can be derived by multiplying the energy in eV with this constant, allowing us to express the energy in both standard scientific and practical units.
  • This high energy is what allows electrons to impact atoms with enough force to affect their inner structure, particularly influencing elements with a high atomic number (Z).
Hydrogenlike Atom
A hydrogenlike atom is an atom that resembles hydrogen in its one-electron nature. Although hydrogen itself has only one proton in the nucleus, the term is extended to other elements with a single electron orbiting around a nucleus with multiple protons. This is an idealized model but very useful in predicting certain physical properties like electron energies in a single-electron atomic system.

In the context of X-ray tube physics, applying the concept of hydrogenlike atoms helps simplify the energy calculations for electrons at various shells. This calculation uses the energy level formula:
  • Energy = \(Z^2\left(-13.6 \, \mathrm{eV} / \mathrm{n}^2\right)\)
Here, \(Z\) is the atomic number, meaning the number of protons in the nucleus, and \(n\) is the principal quantum number representing the electron shell level.
  • For K-shell electrons, \(n = 1\), putting them in close proximity to the nucleus, with high negative potential energy values.
  • Understanding these energy calculations allows us to predict if an electron with a specific amount of kinetic energy can penetrate or eject a K-shell electron in heavier elements.
K-Shell Electron
K-shell electrons reside in the innermost shell of an atom. This shell is closest to the nucleus and is represented by \(n = 1\) in quantum mechanical terms. K-shell electrons in heavier elements play an important role in chemical bonding and atomic behavior, making them significant in several areas of physics and chemistry.
  • These electrons experience the strongest attractive force from the nucleus compared to electrons in other shells.
To knock out a K-shell electron, an incoming electron, such as those in an X-ray tube, must possess energy that matches or exceeds the binding energy of the electron in this shell. This makes electron acceleration calculations essential, as they determine whether the incident electron has enough oomph.
  • For higher atomic number elements, binding energies increase because more protons pull the electrons closer, demanding more energy to detach a K-shell electron.
Understanding this concept simplifies how we interpret X-ray emissions and effects, which are pivotal in medical imaging and material analysis.

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