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Chapter 18: Problem 6

What is the de Broglie wavelength for an electron with speed (a) \(v=0.480 c\) and (b) \(v=0.960 c ?\) (Hint: Use the correct relativistic expression for linear momentum if necessary.)

Short Answer

Expert verified
(a) The de Broglie wavelength is approximately \( 3.55 \times 10^{-12} \; \text{m} \). (b) It is approximately \( 7.06 \times 10^{-13} \; \text{m} \).

Step by step solution

01

Understand the de Broglie Wavelength Formula

The de Broglie wavelength, \( \lambda \), is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{Js} \), and \( p \) is the momentum of the particle.
02

Identify the Expression for Relativistic Momentum

For high-speed particles, such as an electron moving at a significant fraction of the speed of light \( c \), the relativistic momentum \( p \) is given by \[ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \] where \( m \) is the rest mass of the electron, approximately \( 9.109 \times 10^{-31} \, \text{kg} \).
03

Calculate the de Broglie Wavelength for (a) \( v=0.480c \)

First, calculate the relativistic momentum: \[ p_a = \frac{9.109 \times 10^{-31} \times 0.480c}{\sqrt{1 - (0.480)^2}} \]where \( c = 3.00 \times 10^8 \, \text{m/s} \).Compute \( p_a \) and then \( \lambda_a = \frac{6.626 \times 10^{-34}}{p_a} \).
04

Calculate the Specific Values for (a)

Compute the denominator for momentum:\[ \sqrt{1 - (0.480)^2} = \sqrt{1 - 0.2304} = \sqrt{0.7696} \approx 0.877 \]Substitute values to calculate \( p_a \):\[ p_a = \frac{9.109 \times 10^{-31} \times 0.480 \times 3.00 \times 10^8}{0.877} \approx 1.865 \times 10^{-22} \; \text{kg} \cdot \text{m/s} \]Thus, \( \lambda_a = \frac{6.626 \times 10^{-34}}{1.865 \times 10^{-22}} \approx 3.55 \times 10^{-12} \; \text{m} \).
05

Calculate the de Broglie Wavelength for (b) \( v=0.960c \)

Similarly, calculate the relativistic momentum for \( v = 0.960c \):\[ p_b = \frac{9.109 \times 10^{-31} \times 0.960c}{\sqrt{1 - (0.960)^2}} \]Compute \( p_b \) and then \( \lambda_b = \frac{6.626 \times 10^{-34}}{p_b} \).
06

Calculate the Specific Values for (b)

Compute the denominator for momentum:\[ \sqrt{1 - (0.960)^2} = \sqrt{1 - 0.9216} = \sqrt{0.0784} \approx 0.280 \]Substitute values to calculate \( p_b \):\[ p_b = \frac{9.109 \times 10^{-31} \times 0.960 \times 3.00 \times 10^8}{0.280} \approx 9.390 \times 10^{-22} \; \text{kg} \cdot \text{m/s} \]Thus, \( \lambda_b = \frac{6.626 \times 10^{-34}}{9.390 \times 10^{-22}} \approx 7.06 \times 10^{-13} \; \text{m} \).

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Key Concepts

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

Relativistic Momentum
When dealing with particles moving at high speeds, close to the speed of light, the classical formula for momentum doesn't suffice. This is because, at such speeds, the effects of relativity become significant. The relativistic momentum is expressed as:
  • \[ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \]
  • Here, \( m \) is the rest mass of the particle, \( v \) is its velocity, and \( c \) is the speed of light, approximately \( 3.00 \times 10^8 \, \text{m/s} \).
For example, if an electron is moving at speeds like \(0.480c\) or \(0.960c\), where \(c\) is the speed of light, we utilize this formula to calculate its momentum. At speeds approaching the speed of light, the momentum grows significantly due to the influence of the relativistic factor \(\sqrt{1 - \frac{v^2}{c^2}}\), also called the Lorentz factor. This factor affects the momentum calculation by increasing the denominator, hence reducing the overall momentum and allowing accounting for relativistic effects.
Planck's Constant
Planck's constant, denoted as \( h \), is a fundamental physical constant that is a cornerstone of quantum mechanics.
  • It is instrumental in describing the quanta of action in quantum mechanics.
  • The value of Planck's constant is approximately \( 6.626 \times 10^{-34} \, \text{Js} \).
In the context of de Broglie's hypothesis, Planck's constant is used to relate the momentum of a particle to its wavelength through the equation:\[ \lambda = \frac{h}{p} \]This formula shows that the wavelength (\( \lambda \)) of any matter is inversely proportional to its momentum (\( p \)). This relationship is crucial for understanding particle-wave duality, where particles such as electrons exhibit both wave-like and particle-like properties, with Planck's constant linking the quantum and classical worlds.
Electron Rest Mass
The rest mass of an electron is a vital constant often used in calculations involving electrons. It represents the mass of an electron when it is stationary, and it's denoted by \( m_0 \).
  • The rest mass of an electron is approximately \( 9.109 \times 10^{-31} \, \text{kg} \).
When calculating properties like the de Broglie wavelength, the rest mass forms a crucial component of the relativistic momentum expression. Even though the electron's rest mass is tiny, it is central to computations in both quantum mechanics, where the mass is a part of the particle energy, and in relativity, where it combines with the speed of light and velocity to produce relativistic effects. Understanding the rest mass in these contexts aids in grasping why electrons behave uniquely at quantum scales compared to classical physics expectations.

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Most popular questions from this chapter

A hydrogen atom is in a state with energy \(-1.51 \mathrm{eV}\). In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?

A monochromatic light source of frequency \(v\) illuminates a metallic surface and ejects photoelectrons. The photoelectrons emitted are able to excite the hydrogen atom which then emits a photon of wavelength \(1237 \AA\). When the whole experiment is repeated with light source of frequency \(13 v / 10\), the photoelectrons emitted are just able to ionize the hydrogen atom in ground state. Find the work function of the metal and frequency \(v\) of the source.

An electron has a de Broglie wavelength of \(2 \times 10^{-10} \mathrm{~m}\). Determine (a) the magnitude of its momentum and (b) its kinetic energy (in joules and in electron volts).

Zero-Point Energy. Consider a particle with mass \(m\) moving in a potential \(U=\frac{1}{2} k x^{2}\), as in a mass-spring system. The total energy of the particle is \(E=p^{2} / 2 m+\frac{1}{2} k x^{2}\). Assume that \(p\) and \(x\) are approximately related by the Heisenberg uncertainty principle, so \(p x \approx h .\) (a) Calculate the minimum possible value of the energy \(E\), and the value of \(x\) that gives this minimum \(E\). This lowest possible energy, which is not zero, is called the zero-point energy. (b) For the \(x\) calculated in part (a), what is the ratio of the kinetic to the potential energy of the particle?

In an X-ray tube the accelerating voltage is \(20 \mathrm{KV}\). Two targets \(A\) and \(B\) are used one by one. For 'A' the wavelength of the \(K_{\alpha}\) line is \(62 \mathrm{pm}\). For ' \(B\) ' the wavelength of the \(L_{\alpha}\) line is \(124 \mathrm{pm}\). The energy of the ' \(B\) ' ion with vacancy in ' \(L\) ' shell is \(15.5 \mathrm{KeV}\) higher than the atom of \(B\). [Take \(h c=12400 \mathrm{eVA}]\) (i) Find \(\lambda_{\min }\) in \(\hat{A}\). (ii) Can \(K_{\alpha}\) - photon be emitted by 'A'? Explain with reason. (iii) Can \(L\) - photons be emitted by ' \(B\) '? What is the minimum wavelength (in \(A\) ) of the characteristic \(L\) series X-ray that will be emitted by ' \(B\) '.
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