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Calculate, in units of \(\hslash\), the magnitude of the maximum orbital angular momentum for an electron in a hydrogen atom for states with a principal quantum number of 2, 20, and 200. Compare each with the value of n\(\hslash\) postulated in the Bohr model. What trend do you see?

Short Answer

Expert verified
For large \(n\), the quantum and Bohr models converge, showing better classical approximation.

Step by step solution

01

Understanding Maximum Orbital Angular Momentum

In quantum mechanics, the maximum orbital angular momentum for an electron in a given principal quantum number state, denoted as \(n\), is given by \(l_{max} = n - 1\). Angular momentum \(L\) in terms of \(l\) is \(L = \sqrt{l(l+1)}\hslash\).
02

Orbital Angular Momentum for Quantum Number 2

For \(n = 2\), the maximum \(l\) is \(l_{max} = 2 - 1 = 1\). Thus, the maximum orbital angular momentum is: \[ L = \sqrt{1(1 + 1)}\hslash = \sqrt{2}\hslash \]
03

Orbital Angular Momentum for Quantum Number 20

For \(n = 20\), the maximum \(l\) is \(l_{max} = 20 - 1 = 19\). Thus, the maximum orbital momentum is: \[ L = \sqrt{19(19 + 1)}\hslash = \sqrt{380}\hslash \]
04

Orbital Angular Momentum for Quantum Number 200

For \(n = 200\), the maximum \(l\) is \(l_{max} = 200 - 1 = 199\). Thus, the maximum orbital angular momentum is: \[ L = \sqrt{199(199 + 1)}\hslash = \sqrt{39800}\hslash \]
05

Comparison With Bohr Model

In the Bohr model, the angular momentum for each state with principal quantum number \(n\) is assumed to be \(n\hslash\). Compare this for each \(n\):- \(n = 2\): \(2\hslash\) - \(n = 20\): \(20\hslash\) - \(n = 200\): \(200\hslash\). Each compared with quantum mechanical values shows that the quantum mechanic model allows a different prediction due to the factor \(\sqrt{l(l+1)}\), which approaches \(n\hslash\) as \(n\) becomes significantly larger.
06

Observing the Trend

As \(n\) increases, the expression \(\sqrt{l(l+1)}\) approaches \(n\), showing that the quantum mechanical model becomes more similar to the Bohr model for larger \(n\). This indicates that for high quantum numbers, classical approximations hold well.

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Key Concepts

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

Orbital Angular Momentum
In quantum mechanics, **orbital angular momentum** describes the angular motion of particles like electrons within an atom. It's quantized, meaning it takes on discrete values rather than any value. This is different from classical physics where any value is possible. Orbital angular momentum is related to the principal quantum number, \(n\), which enumerates the electron's energy level in an atom, and it depends on another quantum number, \(l\). The maximum value of \(l\) for an electron is \(l_{max} = n - 1\). So, if \(n = 3\), then \(l\) can be 0, 1, or 2. The formula for orbital angular momentum is given by:
  • \(L = \sqrt{l(l+1)}\hslash\)
Here, \(\hslash\) represents the reduced Planck's constant, a fundamental physical constant. This quantization and the use of \(\hslash\) signify the inherent way energy particles take on steps rather than a continuous slide across values—%a hallmark of quantum mechanics.%.
Principal Quantum Number
The principal quantum number, often denoted as \(n\), is a crucial concept in the realm of quantum mechanics. It primarily determines the overall energy level of an electron within an atom. Each integer value of \(n\) corresponds to a different electron shell or energy level.
  • A lower \(n\) value indicates an electron closer to the nucleus, occupying lower energy levels.
  • Higher \(n\) values correspond to electrons that are further from the nucleus, residing in higher energy levels.
These energy levels are quantized, meaning they can only exist in specific, discrete states.The principal quantum number also plays a role in other characteristics of quantum states, such as:- The calculation of radial distance of an electron from the nucleus. - Influencing the size and energy of the electron shell. - Contributing to the complexity of atomic spectra.Understanding this concept helps us grasp how electrons are arranged around an atom and how their configuration influences an atom's chemical behavior.
Bohr Model
The Bohr model, proposed by Niels Bohr in 1913, was one of the earliest models to describe the behavior of electrons within atoms. Despite its simplicity, the Bohr model introduced profound changes to our understanding of atomic structure. In this model, electrons orbit the nucleus much like planets orbit the sun, but with strict rules:
  • The electron's angular momentum is quantized. For each value of the principal quantum number \(n\), angular momentum is \(n\hslash\).
  • Electrons can only occupy certain allowed orbits where their angular momentum is fixed.
  • As electrons move between these orbits, they emit or absorb energy in quantized packets called photons.
While the Bohr model doesn't fully align with current quantum mechanics —it omits electron spin and doesn't accommodate multi-electron systems— it solved certain puzzles of early 20th-century physics and provided a foundation for modern atomic theory.Overall, the Bohr model helped bridge the gap between classical and quantum physics by restricting certain classical principles to specific configurations, aligning surprisingly well proportionally with quantum results for larger quantum states.

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Most popular questions from this chapter

(a) What is the lowest possible energy (in electron volts) of an electron in hydrogen if its orbital angular momentum is \(\sqrt{20}\) \(\hbar$$?\) (b) What are the largest and smallest values of the \(z\)-component of the orbital angular momentum (in terms of \(\hbar\)) for the electron in part (a)? (c) What are the largest and smallest values of the spin angular momentum (in terms of \(\hbar\)) for the electron in part (a)\(?\) (d) What are the largest and smallest values of the orbital angular momentum (in terms of \(\hbar\)) for an electron in the \(M\) shell of hydrogen?

A hydrogen atom undergoes a transition from a 2\(p\) state to the 1\(s\) ground state. In the absence of a magnetic field, the energy of the photon emitted is 122 nm. The atom is then placed in a strong magnetic field in the z-direction. Ignore spin effects; consider only the interaction of the magnetic field with the atom's orbital magnetic moment. (a) How many different photon wavelengths are observed for the 2p \(\rightarrow\) 1s transition? What are the \(m$$_l\) values for the initial and final states for the transition that leads to each photon wavelength? (b) One observed wavelength is exactly the same with the magnetic field as without. What are the initial and final \(m$$_l\) values for the transition that produces a photon of this wavelength? (c) One observed wavelength with the field is longer than the wavelength without the field. What are the initial and final \(m$$_l\) values for the transition that produces a photon of this wavelength? (d) Repeat part (c) for the wavelength that is shorter than the wavelength in the absence of the field.

A particle is described by the normalized wave function \(\psi$$(x, y, z)\) = \(Axe$${^-}{^a}{^x}^2$$e$${^-}{^\beta}$${^y}^2$$e$${^-}{^y}^z$$^2\), where \(A\), \(\alpha\),\(\beta\), and \(\gamma\) are all real, positive constants. The probability that the particle will be found in the infinitesimal volume \(dx\) \(dy\) \(dz\) centered at the point \((x_0\), \(y_0\), \(z_0\)) is \(\mid$$\psi$$(x_0\), \(y_0\), \(z_0\))\(\mid$$^2\) \(dx\) \(dy\) \(dz\). (a) At what value of \(x_0\) is the particle most likely to be found? (b) Are there values of \(x_0\) for which the probability of the particle being found is zero? If so,at what \(x_0$$?\)

An electron is in a three-dimensional box with side lengths \(L_X =\) 0.600 nm and \(L_Y = L_Z = 2L_X\). What are the quantum numbers \(n_X, n_Y,\) and \(n_Z\) and the energies, in eV, for the four lowest energy levels? What is the degeneracy of each (including the degeneracy due to spin)?

The normalized radial wave function for the \(2p\) state of the hydrogen atom is \(R_2{_p}\) = \(( 1/ \sqrt{24a^5}\))\(re$$^-{^r}{^/}{^2}{^a}\). After we average over the angular variables, the radial probability function becomes \(P$$(r)\) \(dr\) = \((R_2{_p}$$)^2\)r\(^2\) \(dr\). At what value of \(r\) is \(P$$(r)\) for the \(2p\) state a maximum? Compare your results to the radius of the \(n\) = 2 state in the Bohr model.

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