Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 34

(a) In terms of the Bohr theory of the hydrogen atom, what process is occurring when excited hydrogen atoms emit radiant energy of certain wavelengths and only those wavelengths? (b) Does a hydrogen atom "expand" or "contract" as it moves from its ground state to an excited state?

Short Answer

Expert verified
(a) When excited hydrogen atoms emit radiant energy of certain wavelengths and only those wavelengths, it represents the process of electron transitions between various energy levels in the hydrogen atom according to the Bohr theory. (b) A hydrogen atom "expands" as it moves from its ground state to an excited state, because the electron transitions to an orbit with a larger size, spending more time farther away from the nucleus.

Step by step solution

01

(Step 1: Understand energy levels in Bohr theory)

(According to Bohr's theory, electrons in a hydrogen atom move in orbits around the nucleus with certain allowed energy levels. When an electron transitions from a higher energy level to a lower energy level, the atom emits light with a specific wavelength. The emitted light corresponds to the energy difference between the two levels.)
02

(Step 2: Relate emitted light to energy transitions)

(The energy difference between the two levels determines the wavelength of the emitted light, according to the following formula: \[\Delta E = E_{final} - E_{initial} = -13.6 \frac{eV}{n^2}\] where \(n\) is the principal quantum number of the energy level, and \(\Delta E\) is the energy difference. The energy difference is also related to the wavelength and frequency of emitted light by the equation: \[\Delta E = h \nu\] where \(h\) is Planck's constant and \(\nu\) is the frequency. Combining the two equations allows us to calculate the wavelength of the emitted light: \[\lambda = \frac{c}{\nu} = \frac{hc}{\Delta E}\] where \(\lambda\) is the wavelength and \(c\) is the speed of light.)
03

(Step 3: Determine change in hydrogen atom size)

(As a hydrogen atom moves from the ground state (n=1) to an excited state (n>1), the electron spends more time farther away from the nucleus because the higher energy orbit is larger in size. This means that the hydrogen atom "expands" as it moves from the ground state to an excited state.)
04

(Answer)

(a) When excited hydrogen atoms emit radiant energy of certain wavelengths and only those wavelengths, it represents the process of electron transitions between various energy levels in the hydrogen atom according to the Bohr theory. (b) A hydrogen atom "expands" as it moves from its ground state to an excited state, because the electron transitions to an orbit with a larger size, spending more time farther away from the nucleus.)

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 hydrogen atom can absorb light of wavelength \(2626 \mathrm{nm}\). (a) In what region of the electromagnetic spectrum is this absorption found? (b) Determine the initial and final values of \(n\) associated with this absorption.

The following electron configurations represent excited states. Identify the element, and write its ground-state condensed electron configuration. (a) \(1 s^{2} 2 s^{2} 3 p^{2} 4 p^{1},\) (b) \([\mathrm{Ar}] 3 d^{10} 4 s^{1} 4 p^{4} 5 s^{1}\), (c) \([\mathrm{Kr}] 4 d^{6} 5 s^{2} 5 p^{1}\) (a) Determine which elements emit radiation in the visible part of the spectrum. (b) Which element emits photons of highest energy? Of lowest energy? (c) When burned, a sample of an unknown substance is found to emit light of frequency \(6.59 \times 10^{14} \mathrm{~s}^{-1}\). Which of these elements is probably in the sample?

Consider a fictitious one-dimensional system with one electron. The wave function for the electron, drawn at the top of the next page, is \(\psi(x)=\sin x\) from \(x=0\) to \(x=2 \pi\). Sketch the probability density, \(\psi^{2}(x),\) from \(x=0\) to \(x=2 \pi\). (b) At what value or values of \(x\) will there be the greatest probability of finding the electron? (c) What is the probability that the electron will be found at \(x=\pi ?\) What is such a point in a wave function called? [Section 6.5\(]\)

Using Heisenberg's uncertainty principle, calculate the uncertainty in the position of (a) a 1.50 -mg mosquito moving at a speed of \(1.40 \mathrm{~m} / \mathrm{s}\) if the speed is known to within \(\pm 0.01 \mathrm{~m} / \mathrm{s} ;\) (b) a proton moving at a speed of \((5.00 \pm 0.01) \times 10^{4} \mathrm{~m} / \mathrm{s}\). (The mass of a proton is given in the table of fundamental constants in the inside cover of the text.)

(a) A red laser pointer emits light with a wavelength of \(650 \mathrm{nm}\). What is the frequency of this light? (b) What is the energy of one of these photons? (c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of \(650 \mathrm{nm}\) photons. What is the energy gap between the ground state and excited state in the laser material?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Physical 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