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The de Broglie equation for a particle can be applied to an electron orbiting a nucleus if one assumes that the electron must have an exact integral number of wavelengths as it covers the circumference of the orbit having radius \(r: n \lambda=2 \pi r\). From this, derive Bohr's quantized angular momentum postulate.

Short Answer

Expert verified
The derivation shows that electron angular momentum is quantized: \( L = \frac{nh}{2\pi} \).

Step by step solution

01

Understand the de Broglie Equation

The de Broglie wavelength formula for a particle is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck's constant and \( p \) is the momentum of the particle. For an electron orbiting a nucleus, its momentum \( p \) is \( mv \), where \( m \) is the mass of the electron and \( v \) is its velocity.
02

Set up the wavelength condition

According to the problem, the electron's wavelength must satisfy \( n\lambda = 2\pi r \), where \( n \) is an integer, \( \lambda \) is the de Broglie wavelength, and \( r \) is the orbit's radius. Substitute the de Broglie wavelength \( \lambda = \frac{h}{mv} \) into this condition, giving us \( n\left(\frac{h}{mv}\right) = 2\pi r \).
03

Simplify the equation

Rearrange the equation from the previous step: \( n\frac{h}{mv} = 2\pi r \) becomes \( nh = 2\pi rmv \). This simplifies to \( nh = 2\pi L \), where \( L = rmv \) is the angular momentum of the electron.
04

Derive Bohr's Quantized Angular Momentum

Set \( L = \frac{nh}{2\pi} \), indicating that the angular momentum \( L \) of the electron is quantized and can only have discrete values defined by \( \frac{nh}{2\pi} \), where \( n \) is an integer. This is Bohr's quantization condition for angular momentum.

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Key Concepts

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

Bohr's quantization condition
Bohr's quantization condition, also known as Bohr's angular momentum quantization, was proposed by Niels Bohr in 1913 as part of his model of the hydrogen atom. According to this condition, an electron can only orbit the nucleus in specific, quantized orbits where its angular momentum is an integer multiple of a fixed unit. This quantization is mathematically represented by:\[ L = \frac{nh}{2\pi} \]In this equation, \(L\) is the electron's angular momentum, \(h\) is Planck's constant, \(n\) is a positive integer known as the quantum number, and \(\pi\) is the mathematical constant pi. This concept was revolutionary at the time because it explained why electrons don't spiral into the nucleus, remaining in stable orbits instead. Bohr's model uses this condition to predict the energy levels of hydrogen and the spectral lines emitted when an electron transitions between these levels.
  • Bohr's condition implies that not all values of angular momentum are possible, but only those where \(n\) is an integer.
  • This was crucial in developing quantum mechanics, as it was one of the first indications that microscopic systems behave very differently from macroscopic ones.
de Broglie wavelength
The concept of the de Broglie wavelength revolutionized the field of quantum mechanics by proposing that every moving particle or object has an associated wave. The de Broglie wavelength \( \lambda \), for a particle with momentum \( p \), is given by the formula:\[ \lambda = \frac{h}{p} \]Here, \(\lambda\) is the wavelength, \(h\) is Planck's constant, and \(p\) is the momentum of the particle. If we consider an electron moving around the nucleus, its momentum \(p\) is the product of its mass \(m\) and velocity \(v\). Therefore, the de Broglie wavelength of an electron becomes \( \lambda = \frac{h}{mv} \).
  • This concept is essential for understanding the dual nature of particles, acting both as particles and waves.
  • For an electron, this wave characteristic allows it to form standing waves in an atom, dictating the orbit's radius, and reinforcing the quantized nature of these systems.
angular momentum quantization
Angular momentum quantization is a fundamental idea in quantum mechanics, where certain physical properties, like angular momentum, can only take on discrete values rather than a continuous range. According to the Bohr model, the angular momentum \(L\) of an electron in an atom's orbit is quantized and is given by:\[ L = mvr = \frac{nh}{2\pi} \]Here, \(m\) is the mass of the electron, \(v\) is its velocity, \(r\) is the radius of the orbit, \(n\) is an integer called the principal quantum number, and \(h\) is Planck's constant. This result stems from combining the de Broglie wavelength condition with the circumference of the orbit.
  • The quantization of angular momentum explains why electrons are found at specific intervals from the nucleus, corresponding to quantized energy levels.
  • This concept helped clarify why atoms emit/absorb energy at specific frequencies, resulting in phenomena such as spectral lines.
  • It underscored the importance of quantization in the microscopic world, laying the groundwork for quantum theory.

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Most popular questions from this chapter

The slope of the plot of energy versus wavelength for the Rayleigh-Jeans law is given by a rearrangement of equation \(9.20\) : $$ \frac{d \rho}{d \lambda}=\frac{8 \pi k T}{\lambda^{4}} $$ What are the value and units of this slope for a blackbody having the following temperatures and at the following wavelengths? (a) \(1000 \mathrm{~K}, 500 \mathrm{~nm}\); (b) \(2000 \mathrm{~K}, 500 \mathrm{~nm}\); (c) \(2000 \mathrm{~K}, 5000 \mathrm{~nm}\); (d) \(2000 \mathrm{~K}, 10,000 \mathrm{~nm}\). Do the answers indicate the presence of an ultraviolet catastrophe?

For an object having mass \(m\) falling in the \(z\) direction, the kinetic energy is \(\frac{1}{2} m z\) and the potential energy is \(m g z\), where \(g\) is the gravitational acceleration constant (approximately \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) ) and \(z\) is the position. For this one-dimensional motion, determine the Lagrangian function \(L\) and write the Lagrangian equation of motion.

(a) Use Wien displacement law to determine the \(\lambda_{\max }\) of the Sun if its surface temperature is \(5800 \mathrm{~K}\). (b) The human eye sees light most efficiently if the light has a wavelength of \(5000 \AA{A}\left(1 \AA{A}=10^{-10} \mathrm{~m}\right)\), which is in the green-blue portion of the spectrum. To what blackbody temperature does that correspond? (c) Compare your answers from the first two parts and comment.

Calculate the energy of a photon having: (a) a wavelength of \(5.42 \times 10^{-6} \mathrm{~m}\); (b) a frequency of \(6.69 \times 10^{13} \mathrm{~s}^{-1}\); (c) a wavelength of \(3.27 \mathrm{~nm}\); (d) a frequency of \(106.5 \mathrm{MHz}\) ( \(1 \mathrm{~Hz}=1\) hertz \(=1 \mathrm{~s}^{-1}\); this unit is often used for frequency); (e) a wavenumber of \(4321 \mathrm{~cm}^{-1}\).

Assume that an electron can absorb more than one photon in the photoelectric effect. (a) How many photons of wavelength \(776.5 \mathrm{~nm}\) does an electron in Fe need to absorb to escape the iron surface \((\phi(\mathrm{Fe})=4.67 \mathrm{eV})\) ? (b) What is the resulting velocity of the emitted electron?

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