/*! 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 17 A wavelength of \(410.2 \mathrm{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 30: Problem 17

A wavelength of \(410.2 \mathrm{nm}\) is emitted by the hydrogen atoms in a high-voltage discharge tube. What are the initial and final values of the quantum number \(n\) for the energy level transition that produces this wavelength?

Short Answer

Expert verified
The initial and final quantum numbers are \( n_1 = 2 \) and \( n_2 = 6 \).

Step by step solution

01

Understanding the Rydberg Formula

The Rydberg formula relates the wavelength of light emitted by an electron transitioning between energy levels in a hydrogen atom. It is given by:\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where \( \lambda \) is the wavelength, \( R_H = 1.097 \times 10^7 \text{ m}^{-1} \) is the Rydberg constant, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final states (with \( n_1 < n_2 \)).
02

Convert Wavelength to Meters

The given wavelength is \(410.2\, \text{nm}\). First, convert this to meters:\[ \lambda = 410.2 \times 10^{-9} \text{ m} \]
03

Apply the Rydberg Formula

Substitute \( \lambda \) and \( R_H \) into the Rydberg formula to find the values of \( n_1 \) and \( n_2 \):\[ \frac{1}{410.2 \times 10^{-9}} = 1.097 \times 10^7 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]Evaluate the left-hand side to calculate \(\frac{1}{\lambda}\).
04

Guess and Check Quantum Numbers

Since \( n_1 \) and \( n_2 \) are integers related to energy levels of hydrogen, start by inputting reasonable values beginning with known transitions. Begin with the Balmer series where \( n_1 = 2 \) and test different values of \( n_2 \) (e.g., 3, 4, 5).
05

Identify Correct Transition

Calculate \( \frac{1}{\lambda} \) for each possible \( n_2 \) and match to the calculated value from Step 3. For \( n_1 = 2 \) and \( n_2 = 6 \), the provided wavelength matches, thus:\[ \frac{1}{410.2 \times 10^{-9}} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{6^2} \right) \] confirms the transition.

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 91影视!

Key Concepts

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

Quantum Numbers
When we talk about quantum numbers, we are discussing the integers that describe the electron's position in an atom. These numbers are crucial because they determine the distribution of an electron around the nucleus.
In the context of hydrogen atoms, the principal quantum number, denoted as \( n \), helps identify the energy level that an electron occupies. It is a positive integer: \( n = 1, 2, 3, \) and so on.
  • In simple terms, the lower the \( n \), the closer the electron is to the nucleus, and the less energy it has.
  • Conversely, a higher \( n \) means the electron is further from the nucleus, with greater energy.
  • These energy levels can also be described as 'shells' around the nucleus.
Quantum numbers don't just describe energy levels. They are integral to understanding how energy transitions occur, as when an electron moves between levels, it emits or absorbs light, as explained by the Rydberg formula.
Balmer Series
The Balmer series is a set of spectral line emissions of the hydrogen atom. Named after Johann Balmer, it is critical in understanding hydrogen's atomic spectrum.
The Balmer series describes transitions of electrons from higher energy levels (with principal quantum number \( n_2 \)) to the second energy level (\( n_1 = 2 \)).
The wavelengths of light emitted during these transitions fall within the visible spectrum.
  • The Balmer series includes distinct spectral lines, such as H-alpha, H-beta, H-gamma, and H-delta.
  • The H-gamma line, for example, corresponds to an electron transition from \( n_2 = 5 \) to \( n_1 = 2 \).
Each line in the series corresponds to a specific energy transition of an electron dropping to the \( n_1 = 2 \) level.
This series is crucial for astronomers as it allows them to identify hydrogen and its abundance in celestial objects through spectroscopy.
Hydrogen Atom Transitions
Within hydrogen atoms, electrons can transition between different energy levels.
These transitions are accompanied by the absorption or emission of light, as described by the Rydberg formula.
The transition in question involves the electron moving between the \( n_2 = 6 \) and \( n_1 = 2 \) energy levels, corresponding to the Balmer series.
  • When the electron drops from a higher energy level (higher \( n_2 \)) to a lower one (lower \( n_1 \)), it emits a photon of light.
  • The energy, and hence the wavelength of this emitted light, is determined by the difference in energy levels.
The calculation of these transitions using the Rydberg formula allows us to predict the spectral lines observed experimentally.
Predicting these transitions precisely helps in understanding not only the hydrogen atom but also provides insights into atomic structures and quantum mechanics in general.

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

There are \(Z\) protons in the nucleus of an atom, where \(Z\) is the atomic number of the element. An \(\alpha\) particle carries a charge of \(+2 \mathrm{e}\). In a scattering experiment, an \(\alpha\) particle, heading directly toward a nucleus in a metal foil, will come to a halt when all the particle's kinetic energy is converted to electric potential energy. In such a situation, how close will an \(\alpha\) particle with a kinetic energy of \(5.0 \times 10^{-13} \mathrm{J}\) come to a gold nucleus \((Z=79) ?\)

In the ground state, the outermost shell \((n=1)\) of helium (He) is filled with electrons, as is the outermost shell \((n=2)\) of neon (Ne). The full outermost shells of these two elements distinguish them as the first two so-called "noble gases." Suppose that the spin quantum number \(m_{\mathrm{s}}\) had three possible values, rather than two. If that were the case, which elements would be (a) the first and (b) the second noble gases? Assume that the possible values for the other three quantum numbers are unchanged, and that the Pauli exclusion principle still applies.

Find the energy (in joules) of the photon that is emitted when the electron in a hydrogen atom undergoes a transition from the \(n=7\) energy level to produce a line in the Paschen series.

A laser is used in eye surgery to weld a detached retina back into place. The wavelength of the laser beam is \(514 \mathrm{nm}\) and the power is 1.5 W. During surgery, the laser beam is turned on for 0.050 s. During this time, how many photons are emitted by the laser?

The electron in a hydrogen atom is in the first excited state, when the electron acquires an additional \(2.86 \mathrm{eV}\) of energy. What is the quantum number \(n\) of the state into which the electron moves?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Electricity

Read Explanation

Thermodynamics

Read Explanation

Quantum Physics

Read Explanation

Measurements

Read Explanation

Space Physics

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.