Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Why does the Bohr model of the hydrogen atom violate the uncertainty principle? (b) In what way is the description of the electron using a wave function consistent with de Broglie's hypothesis? (c) What is meant by the term probability density? Given the wave function, how do we find the probability density at a certain point in space?

Short Answer

Expert verified
(a) Bohr model of the hydrogen atom violates the uncertainty principle because it assumes electrons in fixed circular orbits with defined positions and momentum, which contradicts the principle stating that they cannot be determined simultaneously. (b) Wave function is consistent with de Broglie's hypothesis as it accounts for the electron's wave nature, describing its distribution in space with respect to momentum, instead of using fixed orbits. (c) Probability density describes the likelihood of finding a particle at a certain point in space, and is found by squaring the magnitude of the wave function at that point: \(P(x) = |\psi(x)|^2\).

Step by step solution

01

(a) Bohr model and Uncertainty principle

In the Bohr model of the hydrogen atom, electrons are assumed to orbit the nucleus in certain well-defined circular paths called energy levels. The angular momentum of the electron in each energy level is assumed to be quantized, which means it can have only certain distinct values. The electron has a definite position and momentum in these circular orbits. Meanwhile, the uncertainty principle states that it is impossible to simultaneously determine the exact position and exact momentum of a particle. In the Bohr model, electrons have defined positions (in orbits) and defined momentum since their angular momentum is quantized. This contradicts the uncertainty principle. Therefore, the Bohr model of the hydrogen atom violates the uncertainty principle.
02

(b) Wave function and de Broglie's hypothesis

De Broglie's hypothesis states that particles like electrons have both particle and wave-like properties. This means that particles have an associated wavelength (called de Broglie wavelength) that is inversely proportional to their momentum. Instead of using circular orbits with fixed positions and momentum as in the Bohr model, the wave function describes the electron's behavior in terms of a wave. The wave function is a mathematical function that contains all the information about the electron's position and momentum. In accordance with the de Broglie's hypothesis, the wave function takes into account the wave nature of the electron, and describes its distribution in space with respect to its momentum. This treatment of the electron is consistent with the uncertainty principle and provides a more accurate model of atomic structure than the Bohr model.
03

(c) Probability density definition and calculation

Probability density is a term used to describe the likelihood of finding a particle like an electron at a given point in space. The wave function itself doesn't give the probability directly; instead, it provides the amplitude of the wave, which is a complex number. The probability density at a certain point in space is given by the square of the magnitude of the wave function at that point. Mathematically, it is represented as: \[P(x) = |\psi(x)|^2\] where \(P(x)\) is the probability density at the point x, and \(\psi(x)\) is the wave function at that point. To find the probability density at a certain point in space, we need to square the magnitude of the wave function at that point. This value represents how likely we are to find the particle at that position in space.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The familiar phenomenon of a rainbow results from the diffraction of sunlight through raindrops. (a) Does the wavelength of light increase or decrease as we proceed outward from the innermost band of the rainbow? (b) Does the frequency of light increase or decrease as we proceed outward? (c) Suppose that instead of sunlight, the visible light from a hydrogen discharge tube (Figure 6.10) was used as the light source. What do you think the resulting "hydrogen discharge rainbow" would look like? [Section 6.3]

(a) Consider the following three statements: (i) A hydrogen atom in the \(n=3\) state can emit light at only two specific wavelengths, (ii) a hydrogen atom in the \(n=2\) state is at a lower energy than the \(n=1\) state, and (iii) the energy of an emitted photon equals the energy difference of the two states involved in the emission. Which of these statements is or are true? (b) Does a hydrogen atom "expand" or "contract" as it moves from its ground state to an excited state?

The first 25 years of the twentieth century were momentous for the rapid pace of change in scientists' understanding of the nature of matter. (a) How did Rutherford's experiments on the scattering of \(\alpha\) particles by a gold foil set the stage for Bohr's theory of the hydrogen atom? (b) In what ways is de Broglie's hypothesis, as it applies to electrons, consistent with J. J. Thomson's conclusion that the electron has mass? In what sense is it consistent with proposals preceding Thomson's work that the cathode rays are a wave phenomenon?

The visible emission lines observed by Balmer all involved \(n_{\mathrm{f}}=2\). (a) Which of the following is the best explanation of why the lines with \(n_{\mathrm{f}}=3\) are not observed in the visible portion of the spectrum: (i) Transitions to \(n_{\mathrm{f}}=3\) are not allowed to happen, (ii) transitions to \(n_{\mathrm{f}}=3\) emit photons in the infrared portion of the spectrum, (iii) transitions to \(n_{\mathrm{f}}=3\) emit photons in the ultraviolet portion of the spectrum, or (iv) transitions to \(n_{\mathrm{f}}=3\) emit photons that are at exactly the same wavelengths as those to \(n_{\mathrm{f}}=2\). (b) Calculate the wavelengths of the first three lines in the Balmer series-those for which \(n_{\mathrm{i}}=3,4\), and 5 -and identify these lines in the emission spectrum shown in Figure 6.11.

The Lyman series of emission lines of the hydrogen atom are those for which \(n_{\mathrm{f}}=1\). (a) Determine the region of the electromagnetic spectrum in which the lines of the Lyman series are observed. (b) Calculate the wavelengths of the first three lines in the Lyman series-those for which \(n_{\mathrm{i}}=2,3\), and 4 .

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free