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) What is the smallest amount of energy in electron volts that must be given to a hydrogen atom initially in its ground level so that it can emit the H\(_\alpha\) line in the Balmer series? (b) How many different possibilities of spectral-line emissions are there for this atom when the electron starts in the \(n\) = 3 level and eventually ends up in the ground level? Calculate the wavelength of the emitted photon in each case.

Short Answer

Expert verified
(a) 12.09 eV; (b) 3 transitions, calculate wavelengths using Rydberg formula.

Step by step solution

01

Understanding the Hα Line

The H\(_\alpha\) line is a specific transition in the Balmer series of the hydrogen atom where an electron falls from the n=3 level to the n=2 level. We know that the energy required to excite a hydrogen atom must be enough to allow it to return to the n=3 level from the ground state (n=1).
02

Calculating the Energy for Excitation

The energy of a level in a hydrogen atom is given by the formula: \( E_n = -13.6 \text{ eV} / n^2 \). For n=1 (ground level), \( E_1 = -13.6 \text{ eV} \), and for n=3, \( E_3 \approx -1.51 \text{ eV} \). The energy required to move from n=1 to n=3 is \( E_3 - E_1 = -1.51 \text{ eV} - (-13.6 \text{ eV}) = 12.09 \text{ eV} \). Thus, 12.09 eV is required.
03

Counting Possible Emissions

When in the n=3 state, an electron can transition to n=2, n=1, or first to n=2 and then to n=1. Calculate the number of transitions possible.
04

Calculate Wavelengths of Emissions

For each transition \((n_i \rightarrow n_f)\), use the Rydberg formula: \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \), where \( R_H \) is the Rydberg constant (1.097 × 10^7 m^-1). For 3 to 2, \( n_i = 3, n_f=2 \), and for 3 to 1 and 2 to 1, calculate \( \lambda \). This will yield three wavelengths corresponding to the spectral lines.

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!

Key Concepts

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

Hydrogen Atom
The hydrogen atom is the simplest atom found in the universe, consisting of a single proton and an electron. Understanding this atom is crucial since many foundational concepts of quantum mechanics were developed using the hydrogen atom as a model.
The hydrogen atom's simple structure allows for clear observations and experiments regarding atomic transitions. Being central to quantum mechanics, it exhibits properties explained by the quantum theory, such as distinct energy levels, spectral lines, and behaves according to the principles of quantum physics.
Studying the hydrogen atom provides insights into:
  • Atomic structure
  • Electron behavior in atoms
  • Fundamental principles of energy levels and spectral emissions
So, when we discuss things like the Balmer series, we are looking specifically at what happens to an electron within a hydrogen atom.
Energy Levels
Energy levels in atoms are like the "steps" on which electrons can sit. Each step corresponds to a specific amount of energy. In the hydrogen atom, these energy levels are quantized, meaning that electrons can only occupy certain levels, not those in between.
The energy levels are calculated using the formula:\[E_n = -\frac{13.6 \text{ eV}}{n^2}\]Here:
  • \(E_n\) is the energy of the level,
  • \(n\) is the principal quantum number, an integer indicating the level.
When an electron moves between these levels, it absorbs or emits energy in the form of photons. This leads to spectral lines that we observe in experiments. This concept is pivotal for understanding why specific spectral lines, such as those in the Balmer series, are emitted.
Moving from lower to higher energy levels (like from \(n=1\) to \(n=3\)) requires energy absorption; Conversely, moving back emits energy.
Spectral Lines
Spectral lines are unique signatures of elements, produced when electrons in atoms jump between energy levels. For the hydrogen atom, when electrons transition between levels, they emit light of specific wavelengths, creating the spectral lines we observe.
There are several series of spectral lines, among which the Balmer series is notable. When electrons in a hydrogen atom fall to the second energy level (\(n=2\)), they produce light visible to the human eye, which is part of the Balmer series.
  • Each series corresponds to electrons falling to a different "base" level.
  • Lines in the series have specific colors and wavelengths in the spectrum.
What's fascinating is that these spectral lines serve as a powerful tool to identify elements in distant stars and galaxies, enabling astronomers to understand the composition and dynamics of the universe.
Rydberg Formula
The Rydberg formula is a mathematical expression used to predict the wavelengths of the spectral lines of hydrogen. It is crucial for understanding how and why these lines appear, especially in the Balmer series and others.
The formula is expressed as:\[\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\]where:
  • \(\lambda\) is the wavelength of emitted light,
  • \(R_H\) is the Rydberg constant (1.097 \times 10^7 \text{ m}^{-1}),
  • \(n_i\) is the principal quantum number of the initial energy level,
  • \(n_f\) is the principal quantum number of the final energy level.
Through this formula, we can calculate the wavelengths of light emitted or absorbed by electrons as they move between energy levels. This is especially useful in exercises where we're asked to find the specific wavelengths for transitions within the hydrogen atom, like those involved in producing its spectral lines.

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

A beam of electrons is accelerated from rest and then passes through a pair of identical thin slits that are 1.25 nm apart. You observe that the first double-slit interference dark fringe occurs at \(\pm\)18.0\(^\circ\) from the original direction of the beam when viewed on a distant screen. (a) Are these electrons relativistic? How do you know? (b) Through what potential difference were the electrons accelerated?

What is the de Broglie wavelength for an electron with speed (a) \(v\)= 0.480\(c\) and (b) \(v\) = 0.960\(c\)? (\(Hint\): Use the correct relativistic expression for linear momentum if necessary.)

A hydrogen atom initially in its ground level absorbs a photon, which excites the atom to the \(n\) = 3 level. Determine the wavelength and frequency of the photon.

The shortest visible wavelength is about 400 nm. What is the temperature of an ideal radiator whose spectral emittance peaks at this wavelength?

The radii of atomic nuclei are of the order of 5.0 \(\times\) 10\(^{-15}\) m. (a) Estimate the minimum uncertainty in the momentum of an electron if it is confined within a nucleus. (b) Take this uncertainty in momentum to be an estimate of the magnitude of the momentum. Use the relativistic relationship between energy and momentum, Eq. (37.39), to obtain an estimate of the kinetic energy of an electron confined within a nucleus. (c) Compare the energy calculated in part (b) to the magnitude of the Coulomb potential energy of a proton and an electron separated by 5.0 \(\times\) 10\(^{-15}\) m. On the basis of your result, could there be electrons within the nucleus? (\(Note\): It is interesting to compare this result to that of Problem 39.72.)

See all solutions

Recommended explanations on Physics 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