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Chapter 25: Problem 73

Electrons having de Broglie wavelength \(\lambda\) are incident on a target in a \(X\) -ray tube. Cut-off wavelength of emitted \(X\) -rays is (a) \(\lambda_{o}=\frac{2 m c \lambda^{2}}{h}\) (b) \(\lambda_{o}=\frac{h c}{m c}\) (c) \(\lambda_{o}=\frac{2 m^{2} c^{2} \lambda^{2}}{h^{2}}\) (d) \(\lambda_{o}=0\)

Short Answer

Expert verified
The correct answer is (a) \( \lambda_{o}=\frac{2 m c \lambda^{2}}{h} \).

Step by step solution

01

Understand de Broglie Wavelength

The de Broglie wavelength \( \lambda \) of an electron is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the momentum of the electron. Thus, the momentum of the electron is \( p = \frac{h}{\lambda} \).
02

Calculate Electron's Kinetic Energy

The kinetic energy \( KE \) of an electron can be expressed as \( KE = \frac{p^2}{2m} \), where \( m \) is the mass of the electron. Substituting the expression for \( p \), we have \( KE = \frac{h^2}{2m \lambda^2} \).
03

Determine Cut-off Wavelength Relation

In an X-ray tube, the cut-off wavelength \( \lambda_{o} \) corresponds to the maximum energy of emitted X-rays, which is equal to the kinetic energy of the electron. The energy of the emitted X-ray photon is given by \( E = \frac{hc}{\lambda_{o}} \).
04

Equate Electron's Kinetic and Photon Energies

Equating the kinetic energy of the electron to the energy of an X-ray photon, we have \( \frac{hc}{\lambda_{o}} = \frac{h^2}{2m \lambda^2} \). Solving for \( \lambda_{o} \), we find \( \lambda_{o} = \frac{2m\lambda^2 c}{h} \).
05

Verify the Correct Option

Review the given options to find the one that matches our calculated expression. Option (a) \( \lambda_{o}=\frac{2 m c \lambda^{2}}{h} \) is consistent with our derived expression.

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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 produces X-rays by accelerating electrons at a metal target. When these electrons collide with the target, they suddenly decelerate, and this rapid deceleration causes the emission of X-rays. The tube consists of a cathode, which emits electrons, and an anode, which serves as the metal target. By applying a high voltage between the cathode and the anode, electrons are accelerated and gain kinetic energy.
  • The kinetic energy acquired by the electrons is directly converted into the energy of the emitted X-rays.
  • The intensity and energy of the X-rays can be adjusted by changing the voltage across the tube.
In the operation of an X-ray tube, it's crucial to understand how these parameters affect the characteristics of the resultant X-rays. This understanding is pivotal in many applications, such as medical imaging and material analysis.
cut-off wavelength
The cut-off wavelength in an X-ray tube is a critical concept as it represents the shortest wavelength of X-ray that can be produced. This is the point at which the X-ray maintains its maximum photon energy, directly tied to the kinetic energy of the electrons hitting the anode.
  • Mathematically, the cut-off wavelength \( \lambda_{o} \) is linked to the maximal energy of X-ray produced by the formula: \( E = \frac{hc}{\lambda_{o}} \).
  • The shortest possible wavelength means the highest possible energy of the X-ray photon.
The maximum energy is achieved because all the kinetic energy of the electron is transferred when it collides with the target. Knowing how to calculate the cut-off wavelength is vital for applications requiring high-energy X-rays.
electron kinetic energy
Electron kinetic energy in the context of an X-ray tube is crucial for understanding the X-rays' properties. The kinetic energy of an electron (KE) is dependent on its mass and velocity and is given by the equation \( KE = \frac{p^2}{2m} \), where \( p \) is the momentum of the electron.
  • The momentum itself is defined as \( p = \frac{h}{\lambda} \), with \( \lambda \) being the de Broglie wavelength.
  • As electrons are accelerated in the X-ray tube, they achieve kinetic energy which is largely converted into X-ray energy upon collision.
The conversion of kinetic energy to photon energy underpins the overall energy transformations occurring within an X-ray tube, playing a significant role in the production of X-rays of different intensities.
Planck's constant
Planck's constant, denoted as \( h \), is a fundamental constant in physics that governs the quantization of energy within atomic and subatomic processes. It has a value of approximately \( 6.626 \times 10^{-34} \) Js.
  • Planck's constant relates the energy of a photon to its frequency, given by the equation \( E = h u \), where \( u \) is the frequency.
  • In the context of de Broglie wavelength, it is critical in calculating the momentum of particles: \( p = \frac{h}{\lambda} \).
Understanding Planck's constant is essential for studying wave-particle duality and quantum mechanics, influencing not only theoretical physics but also practical technology such as X-ray generation in medical and industrial fields.

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Most popular questions from this chapter

The maximum velocity of an electron emitted by light of wavelength \(\lambda\) incident on the surface of a metal of work function \(W\) is (a) \(\left[\frac{2(h c+\lambda W)}{m \lambda}\right]^{1 / 2}\) (b) \(\frac{2(h c-\lambda W)}{m}\) (c) \(\left[\frac{2(h c-\lambda W)}{m \lambda}\right]^{1 / 2}\) (d) \(\left[\frac{2(h \lambda-W)}{m}\right]^{1 / 2}\)

The work function for sodium surface is \(2.0 \mathrm{eV}\) and that for aluminium surface is \(4.2 \mathrm{eV}\). The two metals are illuminated with appropriate radiations so as to cause photoemission. Then (a) the threshold frequency for sodium will be less than that for aluminium (b) the threshold frequency of sodium will be more than that of aluminium (c) both sodium and aluminium will have same threshold frequency (d) none of the above

Light of wavelength \(0.6 \mu \mathrm{m}\) from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is \(0.5\) volt. With light of wavelength \(0.40 \mu \mathrm{m}\) from a mercury vapour lamp the stopping potential is \(1.5\) volt; then the work function in electron volts of the photocell surface is (a) \(0.75 \mathrm{eV}\) (b) \(1.5 \mathrm{eV}\) (c) \(3 \mathrm{eV}\) (d) \(2.5 \mathrm{eV}\)

Which one of the following statements is wrong in the context of \(X\) -rays generated from a \(X\) -ray tube? (a) Wavelength of characteristic \(X\) -rays decreases when the atomic number of the target increases (b) Cut-off wavelength of the continuous \(X\) -rays depends on the atomic number of the target (c) Intensity of the characteristic \(X\) -rays depends on the electrical power given to the \(X\) -ray tube (d) Cut-off wavelength of the continuous \(X\) -rays depends on the energy of the electrons in the \(X\) -ray tube

The wavelength of the characteristic \(X\) -ray \(K_{\alpha}\) line emitted by a hydrogen-like element is \(0.32 \AA\). The wave length of \(K_{\beta}\) line emitted by the same element will be (a) \(0.24 \AA\) (b) \(0.27 \AA\) (c) \(0.32 \AA\) (d) \(0.48 \AA\)
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