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Chapter 28: Problem 30

(a) What is the minimum potential difference between the filament and the target of an X-ray tube if the tube is to accelerate electrons to produce X-rays with a wavelength of \(0.150 \mathrm{nm} ?\) (b) What is the shortest wavelength produced in an X-ray tube operated at \(30.0 \mathrm{kV} ?\) (c) Would the answers to parts (a) and (b) be different if the tube accelerated protons instead of electrons? Why or why not?

Short Answer

Expert verified
(a) 8.28 kV, (b) 0.0413 nm, (c) No, answers remain unchanged.

Step by step solution

01

Calculate Energy of X-ray Photon

To find the energy of an X-ray photon with a given wavelength, we use the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{J}\cdot\text{s} \), \( c \) is the speed of light \( 3.00 \times 10^{8} \text{m/s} \), and \( \lambda \) is the wavelength \( 0.150 \text{nm} \) which is \( 0.150 \times 10^{-9} \text{m}\). Calculate \( E \):\[ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{0.150 \times 10^{-9}} \approx 1.325 \times 10^{-15} \text{J} \].
02

Convert Energy to Electronvolts

Convert the energy from joules to electronvolts using the conversion \( 1 \text{eV} = 1.602 \times 10^{-19} \text{J} \). Therefore, \( E \approx \frac{1.325 \times 10^{-15}}{1.602 \times 10^{-19}} \approx 8277 \text{eV} \).
03

Determine Minimum Potential Difference for Part (a)

For the minimum potential difference, we use the energy in electronvolts directly since \( eV = \text{energy} \). Hence \( V \approx 8277 \text{V} \) or approximately \( 8.28 \text{kV} \).
04

Find Shortest Wavelength for Part (b)

The shortest wavelength (i.e., highest energy) X-ray produced is equivalent to the potential difference in energy terms. Using \( V = 30.0 \text{kV} \), the energy is \( E = 30,000 \text{eV} \) and converting to joules gives \( E \approx 4.806 \times 10^{-15} \text{J} \). Apply \( E = \frac{hc}{\lambda} \) to find \( \lambda \): \[ \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{4.806 \times 10^{-15}} \approx 0.0413 \text{nm} \].
05

Assess Impact of Using Protons (Part c)

Protons have a much larger mass than electrons. The kinetic energy gained by charged particles accelerated across a potential difference is the same for electrons and protons, but it affects their velocities differently. For part (c), the minimum potential difference for accelerating protons would still equate to the same energy as electrons, hence the wavelength considerations only rely on energy levels, which remain unchanged.

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Key Concepts

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

X-ray Tubes
X-ray tubes are fascinating devices used to produce X-rays, which are a type of electromagnetic radiation with high energy and short wavelengths. At the heart of an X-ray tube, there is a cathode and an anode. The cathode is a heated filament, which releases electrons when heated. These electrons are then accelerated towards the anode, usually made of tungsten, due to a high potential difference. When the accelerated electrons collide with the target anode, their sudden deceleration leads to the emission of X-rays.
  • The X-rays produced can vary in energy and wavelength, determined by the potential difference used and the target material.
  • To effectively produce X-rays, the electrons must reach very high speeds, which is achieved using a high voltage between the cathode and anode.
Understanding how X-ray tubes work is fundamental to grasp the process of X-ray production and the concepts tied to it, such as photon energy and wavelength.
Photon Energy
Photon energy is a crucial aspect of X-ray production. Photons are particles of light, and their energy can be calculated using the formula: \[E = \frac{hc}{\lambda}\] where \( E \) is the energy of the photon, \( h \) is Planck鈥檚 constant \( (6.626 \times 10^{-34} \text{J}\cdot\text{s}) \), \( c \) is the speed of light \( (3.00 \times 10^8 \text{m/s}) \), and \( \lambda \) is the wavelength.
  • The energy of a photon is inversely proportional to its wavelength; shorter wavelengths have higher energy.
  • In X-ray tubes, the kinetic energy of accelerated electrons is converted into photon energy upon striking the anode.
Being able to calculate photon energy is vital, as it directly relates to the potential difference applied in the X-ray tube and the resulting X-ray wavelengths.
Wavelength Calculation
Calculating the wavelength of X-rays is an essential step in understanding the nature and properties of the produced X-rays. Once we know the photon energy, we can rearrange the energy formula to find the wavelength: \[\lambda = \frac{hc}{E}.\] This calculation helps in determining the shortest or longest wavelength possible from an X-ray tube under specific conditions.
  • In X-ray tubes, the minimum wavelength (maximum photon energy) is linked directly to the highest potential difference used.
  • The calculation of wavelength confirms whether the X-rays produced are useful for imaging or therapeutic purposes.
By understanding how to calculate the wavelength of X-rays, students can predict the outcomes of different setups in X-ray experiments.
Potential Difference
The potential difference in an X-ray tube is the driving force that accelerates electrons from the cathode to the anode. The energy these electrons gain is equivalent to the product of the charge of the electron and the applied potential difference, given by \[E = eV\] where \( e \) is the charge of an electron \( (1.602 \times 10^{-19} \text{C}) \) and \( V \) is the potential difference.
  • A higher potential difference results in higher kinetic energy for the electrons, transforming into higher-energy X-rays upon collision with the anode.
  • The minimum potential difference required is determined based on the desired X-ray wavelength, allowing the precise control of X-ray properties.
Understanding how potential difference influences the production and energy of X-rays is essential for both diagnostic and therapeutic applications in medicine.

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Most popular questions from this chapter

If a photon of wavelength \(0.04250 \mathrm{nm}\) strikes a free electron and is scattered at an angle of \(35.0^{\circ}\) from its original direction, find (a) the change in the wavelength of this photon, (b) the wavelength of the scattered light, (c) the change in energy of the photon (is it a loss or a gain?), and (d) the energy gained by the electron.

An X-ray tube is operated at \(50 \mathrm{kV}\). The shortest wavelength photons from this tube are used in a Compton scattering experiment. One of these photons strikes a free electron and is scattered directly back at an angle of \(180^{\circ} .\) (a) What is the wavelength of the incident photon that comes from the X-ray tube? (b) What is the wavelength of the scattered photon?

Ion microscopes. Just as electron microscopes make use of the wave properties of electrons, ion microscopes use the wave properties of atomic ions, such as helium ions \(\left(\mathrm{He}^{+}\right),\) to image materials. A helium ion has a mass 7300 times that of an electron. In a typical helium ion microscope, helium ions are accelerated by a high voltage of \(10-50 \mathrm{kV}\) and focused on the material to be imaged. At these energies, the ions don't travel very far into the material being imaged, so this type of microscope is used primarily for surface imaging of biological structures. A different method of imaging has been proposed that is sensitive to the entire thickness of the material. This method uses helium ions with much greater energies (in the MeV range), which can pass all the way through biological samples such as cells. In this second type of ion microscope, the energy lost as the ion beam passes through different parts of a cell can be measured and related to the distribution of material in the cell, with thicker parts of the cell causing greater energy loss. How does the wavelength of a helium ion compare to that of an electron accelerated through the same potential difference? A. The helium ion has a longer wavelength because of its greater mass. B. The helium ion has a shorter wavelength because of its greater mass. C. The wavelengths are the same because the kinetic energy is the same. D. The wavelengths are the same because the electric charge is the same.

For a hydrogen atom in the ground state, determine (a) the circumference of the electron orbit, (b) the speed of the electron, (c) the total energy of the electron, and (d) the minimum energy required to remove the electron completely from the atom.

From the kinetic molecular theory of an ideal gas (Chapter 15), we know that the average kinetic energy of an atom is \(\frac{3}{2} k T .\) What is the wavelength of a photon that has this energy for a temperature of \(27^{\circ} \mathrm{C} ?\)
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