Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 152

The wave function for the \(2 s\) orbital in the hydrogen atom is $$ \Psi_{2 s}=\frac{1}{\sqrt{2 a_{0}^{3}}}\left(1-\frac{\rho}{2}\right) e^{-\rho / 2} $$ where \(a_{0}\) is the value of the radius of the first Bohr orbit, equal to \(0.529 \mathrm{nm}, \rho\) is \(Z\left(r / a_{0}\right),\) and \(r\) is the distance from the nucleus in meters. Calculate the location of the node of the \(2 s\) wave function from the nucleus.

Short Answer

Expert verified
The node of the \(2 s\) wave function from the nucleus is at a distance of \(1.058 \times 10^{-9}\) m.

Step by step solution

01

Set the wave function equal to zero

The expression of the wave function is given as \(\Psi_{2 s}=\frac{1}{\sqrt{2 a_{0}^{3}}}(1-\frac{\rho}{2})e^{-\rho/2}\). To find the node, this function should be equal to zero. Therefore, we set \(\Psi_{2 s}\) equal to zero and consider the radial part.
02

Solve for \(\rho\)

When setting \(\Psi_{2 s}\) to zero, note that \(\Psi_{2 s}\) equals zero whenever the expression in the parenthesis equals zero. Therefore, we solve the equation \(1-\frac{\rho}{2} = 0\), which gives \(\rho = 2.0\).
03

Convert \(\rho\) to \(r\)

Now, given that \(\rho = Z(r / a_{0})\) where \(Z = 1\) for hydrogen and that \(a_{0} = 0.529 \cdot 10^{9}\) m, \(r\) can be determined by solving the equation \( r = \rho a_{0} = 2.0 \cdot 0.529 \cdot 10^{-9}\) m.

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

What is the de Broglie wavelength, in centimeters, of a 12.4-g hummingbird flying at \(1.20 \times 10^{2} \mathrm{mph} ?\) \((1 \mathrm{mile}=1.61 \mathrm{~km})\)

The radioactive \(\mathrm{Co}-60\) isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequency of an emitted gamma photon having the energy of \(1.29 \times 10^{11} \mathrm{~J} / \mathrm{mol} .\)

A student placed a large unwrapped chocolate bar in a microwave oven without a rotating glass plate. After turning the oven on for less than a minute, she noticed there were evenly spaced dents (due to melting) about \(6 \mathrm{~cm}\) apart. Based on her observations, calculate the speed of light given that the microwave frequency is \(2.45 \mathrm{GHz}\). (Hint: The energy of a wave is proportional to the square of its amplitude.)

Which quantum number defines a shell? Which quantum numbers define a subshell?

$$ \begin{array}{lccccc} \lambda(\mathrm{nm}) & 405 & 435.8 & 480 & 520 & 577.7 \\ \hline \mathrm{KE}(\mathrm{J}) & 2.360 \times & 2.029 \times & 1.643 \times & 1.417 \times & 1.067 \times \\ & 10^{-19} & 10^{-19} & 10^{-19} & 10^{-19} & 10^{-19} \end{array} $$ (a) What is the lowest possible value of the principal quantum number \((n)\) when the angular momentum quantum number \((\ell)\) is \(1 ?\) (b) What are the possible values of the angular momentum quantum number ( \(\ell\) ) when the magnetic quantum number \(\left(m_{\ell}\right)\) is 0 , given than \(n \leq 4 ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

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