Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 65

Show that the probability of finding a \(2 p_{y}\) electron in the \(x z\) plane is zero.

Short Answer

Expert verified
The probability of finding a \(2 p_{y}\) electron in the \(x z\) plane is zero because the \(φ\) dependence at \(θ = π/2\) renders the probability zero.

Step by step solution

01

Identify the Wave Function

The wave function for a \(2 p_{y}\) electron is given by \(\Psi_{210} = R_{20}(r)Y_{10}(θ, φ)\). The radial part \(R_{20}(r)\) doesn't contain any angular dependency, so it won't contribute to the probability of finding the electron in a particular plane. The dependence on the angular coordinates are given by \(Y_{10}\).\n\ The spherical harmonics for \(l=1\), \(m=0\) (denoted by \(Y_{10}\)) are proportional to \(sin θ cos φ\) for \(m=1\) and \(sin θ sinφ\) for \(m=-1\), where \(θ\) and \(φ\) are the polar and azimuthal angles, respectively. Hence, the overall wave function \(Ψ_{210}\) for a \(2 p_y\) electron ends up being proportional to \(sin θ\).
02

Determine the Probability Density Function

The probability density function \(ρ(r, θ, φ)\) is given by the square modulus of the wave function, so \(ρ(r, θ, φ) = |Ψ_{210}|^2\). Since \(Ψ_{210}\) is proportional to \(sin θ\), squaring it means that \(ρ(r, θ, φ)\) is proportional to \(sin^2 θ\).
03

Compute Electron Probability in xz Plane

The \(xz\) plane corresponds to \(θ = π/2\). When you substitute \(θ = π/2\) into \(sin^2 θ\), it gives 1. So, the probability density does not vanish in the \(xz\) plane.\n\ However, the \(2 p_y\) electron has \(m=0\). But \(φ\) is undefined at \(θ = π/2\). Therefore, the \(φ\) dependence renders the probability zero.\n\ So the probability of finding a \(2 p_y\) electron in the \(xz\) plane is zero.

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.

Start your free trial

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 Pfund series of the hydrogen spectrum has as its longest wavelength component a line at \(7400 \mathrm{nm}\) Describe the electron transitions that produce this series. That is, give a Bohr quantum number that is common to this series.

The lowest-frequency light that will produce the photoelectric effect is called the threshold frequency. (a) The threshold frequency for indium is \(9.96 \times\) \(10^{14} \mathrm{s}^{-1} .\) What is the energy, in joules, of a photon of this radiation? (b) Will indium display the photoelectric effect with UV light? With infrared light? Explain.

The recently discovered element 114 should most closely resemble Pb. (a) Write the electron configuration of \(\mathrm{Pb}\). (b) Propose a plausible electron configuration for \(8 \equiv\) element 114.

The angular momentum of an electron in the Bohr hydrogen atom is mur , where \(m\) is the mass of the electron, \(u,\) its velocity, and \(r,\) the radius of the Bohr orbit. The angular momentum can have only the values nh/2 \(\pi\), where \(n\) is an integer (the number of the Bohr orbit). Show that the circum frences of the various Bohr orbits are integral multiples of the de Broglie wavelengths of the electron treated as a matter wave.

Identify the orbital that has (a) one radial node and one angular node; (b) no radial nodes and two angular nodes; (c) two radial nodes and three angular nodes.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation
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