/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Indicate whether energy is emitt... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Indicate whether energy is emitted or absorbed when the following electronic transitions occur in hydrogen: (a) from \(n=2\) to \(n=6,\) (b) from an orbit of radius \(4.76 \AA\) to one of radius \(0.529 \AA,(\mathrm{c})\) from the \(n=6\) to the \(n=9\) state.

Short Answer

Expert verified
In the given electronic transitions, energy is absorbed when transitioning from n=2 to n=6 and from n=6 to n=9. Energy is emitted when transitioning from an orbit of radius 4.76 Ã… to one of radius 0.529 Ã….

Step by step solution

01

Compare Initial and Final Principal Quantum Numbers

In this case, the electron transitions from n=2 to n=6. Since the final principal quantum number (n=6) is greater than the initial (n=2), the electron moves to a higher energy orbit.
02

Determine If Energy is Emitted or Absorbed

When moving to a higher energy orbit, the electron absorbs energy. Therefore, during the transition from n=2 to n=6, energy is absorbed. #b)_Transition_from_4.76_Ã…_to_0.529_Ã…#
03

Relate Orbital Radius to Principal Quantum Number

The orbital radius (r) of hydrogen is related to the principal quantum number (n) by the formula: \(r = n^2a_0\), where \(a_0\) is the Bohr radius (approximately 0.529 Ã…).
04

Calculate Initial and Final Principal Quantum Numbers

Given the initial radius of 4.76 Ã… and the final radius of 0.529 Ã…, we can find the initial and final principal quantum numbers by rearranging the equation: Initial: \(n_1 = \sqrt{r_1/a_0} = \sqrt{4.76\text{ Ã…} / 0.529\text{ Ã…}} \approx 3\) Final: \(n_2 = \sqrt{r_2/a_0} = \sqrt{0.529\text{ Ã…} / 0.529\text{ Ã…}} = 1\)
05

Determine If Energy is Emitted or Absorbed

Since the final principal quantum number (n=1) is lower than the initial (n=3), the electron moves to a lower energy orbit. Therefore, during the transition from a radius of 4.76 Ã… to 0.529 Ã…, energy is emitted. #c)_Transition_from_n=6_to_n=9#
06

Compare Initial and Final Principal Quantum Numbers

In this case, the electron transitions from n=6 to n=9. Since the final principal quantum number (n=9) is greater than the initial (n=6), the electron moves to a higher energy orbit.
07

Determine If Energy is Emitted or Absorbed

When moving to a higher energy orbit, the electron absorbs energy. Therefore, during the transition from n=6 to n=9, energy is absorbed. In conclusion, for the given electronic transitions: a) Energy is absorbed. b) Energy is emitted. c) Energy is absorbed.

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 91Ó°ÊÓ!

Key Concepts

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

Principal Quantum Number
The principal quantum number, denoted as n, is a fundamental concept in quantum mechanics which describes the size and energy level of an electron's orbit in an atom. Think of it as the 'address' of the electron within the atomic structure. Higher values of n correspond to larger orbits and hence electrons further away from the nucleus, with increased energy. The further an electron is from the nucleus, the more energy it has.

When looking at electronic transitions, such as an electron moving from n=2 to n=6, the principal quantum number increases. This means the electron is absorbing energy to move to a higher energy state. Alternatively, when an electron falls to a lower energy state, like moving from n=3 to n=1, the principal quantum number decreases and the energy is released in the form of electromagnetic radiation, typically observed as light.
Bohr Model of the Hydrogen Atom
The Bohr model of the hydrogen atom, proposed by Niels Bohr in 1913, is a simple, yet powerful way to visualize atomic structure and electron transitions within hydrogen - the most simple atom. This model posits that electrons orbit the nucleus in distinct paths called orbits or shells with fixed sizes and energies, very much like planets around the Sun.

In the context of the Bohr model, the energy levels are quantized, meaning the electron can only exist in certain allowed orbits, corresponding to certain principal quantum numbers. It uses the Bohr radius a0 as the unit for these orbit sizes, which correspond to the lowest energy state, n=1. Transitions between these fixed orbits involve discrete changes in energy, either absorbed when moving to larger orbits or emitted when moving to smaller ones. This discrete nature explains the characteristic emission or absorption spectra observed for hydrogen.
Energy Absorption and Emission
The process of energy absorption and emission in an atom happens when an electron makes a transition between different energy levels or orbits. When an electron jumps to a higher energy level, it absorbs a specific amount of energy, which matches the energy difference between the initial and final levels. Conversely, an electron emits energy when it falls to a lower energy level.

Determining whether energy is absorbed or emitted can be deduced by comparing the principal quantum numbers of an electron’s initial and final orbits. An increase from n=2 to n=6, or n=6 to n=9, signifies absorption. A decrease, from an orbit of radius 4.76 Å to 0.529 Å, which correspond to n=3 to n=1, indicates emission. The emitted or absorbed energy can manifest as different forms of electromagnetic radiation, including visible light, with the exact frequency of the radiation depending on the energy difference between the electron's initial and final state.

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 discovery of hafnium, element number \(72,\) provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements \(58-71\) ) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), is reacted with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and CO in the ratio 1: 2 . Write a balanced chemical equation for the reaction. Starting with a 55.4-g sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\)

Neutron diffraction is an important technique for determining the structures of molecules. Calculate the velocity of a neutron needed to achieve a wavelength of \(0.955 \AA .\) (Refer to the inside cover for the mass of the neutron).

The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of \(8.95 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What is the characteristic wavelength of this electron? Is the wavelength comparable to the size of atoms?

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.

(a) What is the relationship between the wavelength and the frequency of radiant energy? (b) Ozone in the upper atmosphere absorbs energy in the \(210-230-\mathrm{nm}\) range of the spectrum. In what region of the electromagnetic spectrum does this radiation occur?

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.