/*! 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 31 Calculate the wavelength (in \(\... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Calculate the wavelength (in \(\mathrm{nm}\) ) of a photon emitted by a hydrogen atom when its electron drops from the \(n=5\) state to the \(n=3\) state.

Short Answer

Expert verified
The wavelength of the photon emitted when an electron drops from the \(n=5\) state to the \(n=3\) state in a hydrogen atom is approximately 1281 nm.

Step by step solution

01

Identify the energy levels

First, identify the initial 'n initial' and final 'n final' energy levels. From the problem, \(n_{initial} = 5\) and \(n_{final} = 3\).
02

Apply Rydberg’s Formula

Next, apply Rydberg’s formula which is \[\lambda = \frac{1}{R} \left | \frac{1}{n_{final}^{2}} - \frac{1}{n_{initial}^{2}} \right |\] where R is the Rydberg constant \(\approx 1.097373 *10^{7} \, m^{-1}\)
03

Substitute the values

Substitute the values into the Rydberg equation which gives \[\lambda = \frac{1}{1.097373 *10^{7}\, m^{-1}}\left | \frac{1}{3^{2}} - \frac{1}{5^{2}}\right |\]
04

Calculate the result

After calculating the above expression, the wavelength obtained is in meters. Convert it to nanometers (1 meter = \(1.0*10^{9}\) nm). The result should be to 3 significant figures.

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.

Wavelength Calculation
Wavelength calculation is an essential concept in understanding the behavior of light and other forms of electromagnetic radiation as they move through space. For instance, when studying the hydrogen emission spectrum, we frequently calculate the wavelength of light emitted as electrons transition between different energy levels within an atom.

Take the hydrogen atom, for example. When an electron falls from a higher energy level to a lower one, it emits a photon, and this photon carries energy that can be described in terms of wavelength. To calculate this wavelength, we use the Rydberg formula: \[ \lambda = \frac{1}{R} \left | \frac{1}{n_{final}^{2}} - \frac{1}{n_{initial}^{2}} \right | \] where \( R \) is the Rydberg constant and \( n_{initial} \) and \( n_{final} \) are the quantum numbers of the initial and final energy levels.

The units used in these calculations are also important. Typically, wavelengths are measured in meters, but for convenience and due to the small size of the values involved in atomic transitions, we generally convert them to nanometers (nm) where 1 nm = \( 1.0 \times 10^{-9} \) meters. This conversion is vital to express the result in a more readable and practical form.
Hydrogen Emission Spectrum
The hydrogen emission spectrum is a series of specific wavelengths of light that hydrogen atoms emit when their electrons transition from higher to lower energy levels. This spectrum serves as a crucial piece of evidence for the quantized nature of energy in atoms.

The visible part of this spectrum consists of lines known as the Balmer series, and it arises due to the electrons falling to the second energy level (\( n=2 \)) from higher levels. Each of these transitions corresponds to a specific wavelength of light, producing the characteristic lines we can observe.

Understanding Emission Lines

Each line in the hydrogen emission spectrum originates from a specific electronic transition between energy levels. The calculation of the emitted photon's wavelength, as addressed in the exercise, allows us to predict the position and color of these lines in the spectrum.
Quantum Energy Levels
Quantum energy levels are discrete values of energy that an electron within an atom can have. According to quantum mechanics, electrons cannot occupy an 'in-between' state; they must reside in specific layers, or 'shells,' each having a unique energy value.

In the context of a hydrogen atom, these energy levels are determined by the principal quantum number \( n \). The higher the value of \( n \), the farther the electron is from the nucleus and the higher its energy. When an electron transitions between these levels, it either absorbs or emits energy in discrete quantities.

Energy Absorption and Emission

For an electron to move to a higher energy level (\
Photon Emission
Photon emission occurs when an electron within an atom transitions from a higher energy level to a lower one, releasing energy in the form of a photon. This photon carries away energy as electromagnetic radiation and its characteristics, such as frequency and wavelength, are directly correlated to the energy difference between the two levels.

The energy of a single photon can be expressed using Planck's equation: \( E = h u \), where \( E \) is the energy, \( h \) is Planck's constant, and \( u \) is the frequency of the photon. Since the frequency \( u \) is related to the wavelength \( \lambda \) by the equation \( c = \lambda u \) (where \( c \) is the speed of light), we can see how the wavelength of the emitted photon is inversely proportional to the energy difference it represents.

In the exercise context, when an electron drops from the \( n=5 \) state to the \( n=3 \) state, it emits a photon with a specific wavelength calculable using the Rydberg formula. This emitted radiation is just one quanta of light in the hydrogen emission spectrum, which can be observed and measured.

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 sun is surrounded by a white circle of gaseous material called the corona, which becomes visible during a total eclipse of the sun. The temperature of the corona is in the millions of degrees Celsius, which is high enough to break up molecules and remove some or all of the electrons from atoms. One way astronomers have been able to estimate the temperature of the corona is by studying the emission lines of ions of certain elements. For example, the emission spectrum of \(\mathrm{Fe}^{14+}\) ions has been recorded and analyzed. Knowing that it takes \(3.5 \times 10^{4} \mathrm{~kJ} / \mathrm{mol}\) to convert \(\mathrm{Fe}^{13+}\) to \(\mathrm{Fe}^{14+}\), estimate the temperature of the sun's corona. (Hint: The average kinetic energy of one mole of a gas is \(\left.\frac{3}{2} R T .\right)\)

Calculate the total number of electrons that can occupy (a) one \(s\) orbital, (b) three \(p\) orbitals, (c) five \(d\) orbitals, (d) seven \(f\) orbitals.

A photoelectric experiment was performed by separately shining a laser at \(450 \mathrm{nm}\) (blue light) and a laser at \(560 \mathrm{nm}\) (yellow light) on a clean metal surface and measuring the number and kinetic energy of the ejected electrons. Which light would generate more electrons? Which light would eject electrons with greater kinetic energy? Assume that the same amount of energy is delivered to the metal surface by each laser and that the frequencies of the laser lights exceed the threshold frequency.

Draw the shapes (boundary surfaces) of the following orbitals: (a) \(2 p_{y},\) (b) \(3 d_{z^{2}}\) (c) \(3 d_{x^{2}-y^{2}}\), (Show coordinate axes in your sketches.)

List all the possible subshells and orbitals associated with the principal quantum number \(n,\) if \(n=5\)

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.