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Consider an X-ray tube that uses platinum \((Z=78)\) as its target. (a) Use the Bohr model to estimate the minimum kinetic energy electrons must have in order for \(\mathrm{K}_{\alpha}\) X-rays to just appear in the X-ray spectrum of the tube. (b) Assuming the electrons are accelerated from rest through a voltage \(V\), estimate the minimum voltage necessary to produce the \(\mathrm{K}_{a} \mathrm{X}\) -rays.

Short Answer

Expert verified
Minimum electron kinetic energy is 62,198 eV; minimum voltage is 62,198 V.

Step by step solution

01

Understanding the Bohr Model

The Bohr model is used to calculate the energy levels of electrons in an atom. For X-ray emissions, we consider electrons transitioning between energy levels, specifically between the K-shell and L-shell for \( \mathrm{K}_{\alpha} \) X-rays. The energy of the emitted X-ray photon is equal to the difference in energy between these shells.
02

Determine the Energy of the K and L Shells

The energy of a shell can be calculated using the equation \( E_n = -13.6 \frac{Z^2}{n^2} \, \text{eV} \), where \( Z \) is the atomic number and \( n \) is the principal quantum number of the shell. For the K-shell (n=1) and L-shell (n=2) of platinum \( Z=78 \), calculate both energies:
03

Calculate K-shell Energy

For the K-shell (n=1): \[ E_1 = -13.6 \frac{78^2}{1^2} = - 82,930.4 \, \text{eV} \]
04

Calculate L-shell Energy

For the L-shell (n=2): \[ E_2 = -13.6 \frac{78^2}{2^2} = - 20,732.6 \, \text{eV} \]
05

Compute Energy Difference for \( \mathrm{K}_{\alpha} \) X-rays

The energy of the emitted \( \mathrm{K}_{\alpha} \) X-ray is the difference: \[ \Delta E = E_2 - E_1 = -20,732.6 - (-82,930.4) = 62,197.8 \, \text{eV} \]
06

Calculate Minimum Kinetic Energy

The minimum kinetic energy of the electrons must be equal to the energy of the emitted \( \mathrm{K}_{\alpha} \) X-rays. Thus, the minimum kinetic energy required is approximately 62,198 eV.
07

Determine Minimum Voltage

Since the electrons are accelerated from rest through a voltage \( V \), the kinetic energy \( eV \) is equal to the energy of the \( \mathrm{K}_{\alpha} \) X-rays. Therefore, \( V \) must be at least \( 62,198 \, \text{V} \).

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Key Concepts

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

X-ray tube
An X-ray tube is a device that generates X-rays by accelerating electrons at high speeds to collide with a metal target. The basic components of an X-ray tube include:
  • A cathode, which is a heated filament that emits electrons.
  • An anode, which serves as the metal target; in this context, platinum is used.
  • A vacuum tube, which allows electrons to move freely from the cathode to the anode.
When electrons hit the target, their sudden deceleration leads to the emission of X-rays.
These X-rays are produced through two main processes: Bremsstrahlung (braking radiation) and characteristic X-ray production.
In Bremsstrahlung, X-rays result from the deceleration of electrons, while characteristic X-rays result from transitions between energy levels within atoms of the target material.
X-ray tubes are widely used in medical imaging, material analysis, and in this exercise, the generation of K-alpha X-rays provides a valuable insight into the atomic structure.
K-alpha X-rays
K-alpha X-rays are a specific type of characteristic X-rays emitted when an electron from a higher energy level fills a vacancy in the K-shell, which is the innermost electron shell of an atom.
The transition most commonly occurs from the L-shell (n=2) to the K-shell (n=1). This results in the emission of X-ray photons with energies characteristic of the target material.
In an X-ray tube, when electrons have sufficient kinetic energy to knock out electrons from the K-shell, this vacancy is often filled by an electron from the L-shell.
The energy released in this process manifests as a K-alpha X-ray. The equation used to determine the energy of these photons is derived from the Bohr model:
  • The energy of an electron in a shell is given by \( E_n = -13.6 \frac{Z^2}{n^2} \, \text{eV} \), where \( Z \) is the atomic number.
  • The energy of the K-shell (n=1) and L-shell (n=2) can be calculated for a given element, such as platinum with \( Z=78 \).
By calculating the difference in energy between these shells, we derive the energy of the K-alpha X-rays.
Energy levels
In the context of X-ray tubes and atomic physics, energy levels refer to the quantized states that electrons can occupy within an atom.
The Bohr model introduces the concept of electrons moving in discrete orbits around the nucleus, with each orbit corresponding to a different energy level.
  • The principal quantum number, \( n \), determines the size and energy of the orbit.Energy levels are calculated using the formula \( E_n = -13.6 \frac{Z^2}{n^2} \, \text{eV} \), which shows dependency on the atomic number \( Z \) and \( n \).
  • Electrons naturally seek the lowest possible energy state; however, they can be excited to higher levels by absorbing energy.
  • When electrons transition between these levels, they emit or absorb photons, producing characteristic X-ray radiation like K-alpha X-rays.
These energy levels define not just the spectral lines observed but also guide practical applications such as the design of techniques to study and utilize characteristic X-rays.

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