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Chapter 39: Problem 40

(I) Show that the cutoff wavelength \(\lambda_{0}\) in an \(X\) -ray spectrum is given by $$\lambda_{0}=\frac{1240}{V} \mathrm{nm}$$ where \(V\) is the \(X\) -ray tube voltage in volts.

Short Answer

Expert verified
The cutoff wavelength \( \lambda_0 \) in an X-ray spectrum is indeed \( \lambda_{0} = \frac{1240}{V} \mathrm{nm} \), showing a direct relationship between wavelength and tube voltage.

Step by step solution

01

Understand the Bremsstrahlung Equation

The cutoff wavelength in an X-ray spectrum is related to the maximum photon energy, which occurs during bremsstrahlung when all the kinetic energy of an electron is converted into a single X-ray photon. This energy can be expressed as: \( E = h u = h\frac{c}{\lambda_0} \), where \( E \) is the energy of the photon, \( h \) is Planck's constant, \( u \) is the frequency, and \( c \) is the speed of light in vacuum.
02

Connect Energy and Voltage

The kinetic energy of the electron, which is equal to the energy of the emitted X-ray photon at maximum energy, is \( E = eV \), where \( e \) is the charge of the electron, and \( V \) is the voltage of the X-ray tube. Thus, substituting the expression for energy, we have: \( eV = h\frac{c}{\lambda_0} \).
03

Rearrange to Solve for \( \lambda_0 \)

Rearrange the equation \( eV = h\frac{c}{\lambda_0} \) to solve for \( \lambda_0 \):\[ \lambda_0 = \frac{hc}{eV} \]
04

Substitute Constants

Substitute the known values for the constants into the equation: - \( h = 6.626 \times 10^{-34} \mathrm{Js} \),- \( c = 3 \times 10^8 \mathrm{m/s} \), and - \( e = 1.6 \times 10^{-19} \mathrm{C} \). Using these values, you get: \[ \lambda_0 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{(1.6 \times 10^{-19})V} \]
05

Simplify the Expression

Calculate the numerator: \( 6.626 \times 10^{-34} \times 3 \times 10^8 = 1.9878 \times 10^{-25} \mathrm{Js} \), and then divide this by \( 1.6 \times 10^{-19} \) to simplify:\( \frac{1.9878 \times 10^{-25}}{1.6 \times 10^{-19}} = 1.242375 \times 10^{-6} \mathrm{m/V} \). Convert this to nanometers (1 nm = \( 10^{-9} \) m), so:\( 1.242375 \times 10^{-6} \mathrm{m/V} = 1242.375 \mathrm{nm/V} \). This is approximately \( \lambda_0 = \frac{1240}{V} \mathrm{nm} \), which matches the given expression by rounding to three significant figures.

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Key Concepts

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

X-ray tube voltage
X-ray tube voltage is a critical factor in determining the properties of X-rays produced by an X-ray machine. This voltage, usually measured in kilovolts (kV), accelerates electrons towards a target anode. The higher the voltage, the greater the energy these electrons will have upon impact. When these high-energy electrons strike the anode, they decelerate rapidly, leading to the emission of X-rays.
  • The energy of the X-rays generated is directly proportional to the tube voltage.
  • Higher voltage results in shorter wavelength X-rays, which are more penetrating.
Understanding the relationship between tube voltage and X-ray properties is crucial for medical imaging and industrial applications. It allows for the optimization of image quality while minimizing exposure risks.
Bremsstrahlung
Bremsstrahlung, a German term meaning "braking radiation," describes the emission of X-rays when high-energy electrons are decelerated or "braked" by the electric field of a nucleus in the X-ray tube's target material. This interaction is responsible for the majority of X-rays produced in an X-ray tube.
  • When electrons are slowed down, they lose energy, which is emitted as X-ray photons.
  • The spectrum of Bremsstrahlung is continuous, meaning it produces X-rays with a range of energies and wavelengths.
This phenomenon is essential for generating X-rays, providing the continuous spectrum that forms the basis of the X-ray images used in diagnostics and research.
Photon energy
Photon energy is a measure of the energy carried by a single photon, the basic unit of light and all other forms of electromagnetic radiation like X-rays. For X-rays, high-energy photons are produced due to the high velocities of electrons impacting the target material.
  • The energy of a photon is calculated as: \[ E = h \cdot f \] where \( E \) is energy, \( h \) is Planck's constant, and \( f \) is the frequency of the light.
In the context of X-rays, photon energy is crucial in determining the penetration capability of the X-rays. Higher energy photons (shorter wavelength) are more capable of penetrating materials, which is important in medical and industrial imaging to visualize internal structures.
Planck's constant
Planck's constant, denoted as \( h \), is a fundamental constant of nature that plays a crucial role in quantum mechanics. It has a value of approximately \( 6.626 \times 10^{-34} \text{Js} \). This constant is pivotal in the Planck-Einstein relation, which connects the energy of a photon with its frequency: \[ E = h \cdot f \]
  • This relationship is fundamental in understanding processes like photoelectric effects and bremsstrahlung where energy and electromagnetic waves interact.
  • Planck's constant provides the basis for calculating photon energy, a key factor in the production and characteristics of X-rays.
Planck's constant underscores the quantized nature of energy and is integral to defining the boundary between classical and quantum physics.

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Most popular questions from this chapter

(II) Silver atoms \(\left(\right.\) spin \(\left.=\frac{1}{2}\right)\) are placed in a 1.0-T magnetic field which splits the ground state into two close levels. (a) What is the difference in energy between these two levels, and \((b)\) what wavelength photon could cause a transition from the lower level to the upper one? (c) How would your answer differ if the atoms were hydrogen?

(II) Show that a population inversion for two levels (as in a pumped laser) corresponds to a negative Kelvin temperature in the Boltzmann distribution. Explain why such a situation does not contradict the idea that negative Kelvin temperatures cannot be reached in the normal sense of temperature.

For each of the following atomic transitions, state whether the transition is allowed or forbidden, and why: \((a) 4 p \rightarrow 3 p\); (b) \(3 p \rightarrow 1 s ;(c) 4 d \rightarrow 3 d ;(d) 4 d \rightarrow 3 s\) (e) \(4 s \rightarrow 2 p\).

(I) List the quantum numbers for each electron in the ground state of \((a)\) carbon \((Z=6)\) (b) aluminum \((Z=13)\).

(I) If the quantum state of an electron is specified by \(\left(n, \ell, m_{\ell}, m_{s}\right),\) estimate the energy difference between the states \(\left(1,0,0,-\frac{1}{2}\right)\) and \(\left(1,0,0,+\frac{1}{2}\right)\) of an electron in the \(1 s\) state of helium in an external magnetic field of \(2.5 \mathrm{~T}\)
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