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From the energy-level diagram for hydrogen, explain the observation that the frequency of the second Lyman-series line is the sum of the frequencies of the first Lyman-series line and the first Balmer-series line. This is an example of the empirically discovered Ritz combination principle. Use the diagram to find some other valid combinations.

Short Answer

Expert verified
The frequency of the second Lyman line is the sum of the first Lyman and first Balmer series lines, as verified by the Rydberg formula and Ritz combination principle.

Step by step solution

01

Understand the Lyman and Balmer Series

The Lyman series consists of transitions where electrons fall to the n=1 energy level from higher levels (e.g., n=2, n=3, etc.). The Balmer series involves transitions where electrons fall to the n=2 level from higher levels (e.g., n=3, n=4, etc.) in the hydrogen atom.
02

Identify Relevant Transitions

For the second Lyman-series line, we consider the transition from n=3 to n=1. For the first Lyman-series line, consider the transition from n=2 to n=1. For the first Balmer-series line, consider n=3 to n=2.
03

Calculate Frequencies Using Rydberg Formula

The frequency of a transition can be calculated using the Rydberg formula: \[ u = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where \( u \) is frequency, \( R_H \) is the Rydberg constant, \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy levels, respectively.
04

Calculate Frequency of Second Lyman-Series Line

For the second Lyman-line transition (n=3 to n=1):\[ u_3-1 = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R_H \left(1 - \frac{1}{9}\right) = R_H \times \frac{8}{9} \]
05

Calculate Frequency of First Lyman-Series Line

For the first Lyman-line transition (n=2 to n=1):\[ u_2-1 = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left(1 - \frac{1}{4}\right) = R_H \times \frac{3}{4} \]
06

Calculate Frequency of First Balmer-Series Line

For the first Balmer-line transition (n=3 to n=2):\[ u_3-2 = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left(\frac{1}{4} - \frac{1}{9}\right) = R_H \times \frac{5}{36} \]
07

Check Frequency Combination According to Ritz Combination Principle

We check if the frequency of the second Lyman-series line equals the sum of the frequencies of the first Lyman-series line and the first Balmer-series line:\[ u_3-1 = u_2-1 + u_3-2 \]Using the previously calculated frequencies:\[ R_H \times \frac{8}{9} = R_H \times \frac{3}{4} + R_H \times \frac{5}{36} \]Simplifying gives:\[ R_H \times \frac{8}{9} = R_H \times \frac{27}{36} + R_H \times \frac{5}{36} = R_H \times \frac{32}{36} = R_H \times \frac{8}{9} \]

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Key Concepts

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

Lyman Series
The Lyman series is a particular sequence of spectral lines in the hydrogen atom. It occurs when an electron transitions from a higher energy level down to the n=1 level, also known as the ground state. This series includes several possible transitions: an electron can move from n=2 to n=1, from n=3 to n=1, and so forth.
Each transition releases a photon with a distinct frequency and energy. The Lyman series is noted for its ultraviolet (UV) emissions, which are invisible to the naked eye. However, these lines are crucial in studying astronomical bodies and analyzing atomic spectra.

Key points about the Lyman series:
  • Involves transitions to the ground state (n=1).
  • Associated with high-energy UV emissions.
  • Flows from the second level upward to the first level (n=2 or n=3 down to n=1).
Understanding where these transitions occur helps decipher the energy changes within an atom and connect these changes with observable phenomena like spectra.
Balmer Series
The Balmer series also refers to spectral lines from electron transitions within a hydrogen atom, but this series deals specifically with transitions ending at the n=2 level. An electron may drop from levels like n=3, n=4, etc., down to n=2. This series is particularly famous because it appears in the visible light spectrum, enabling it to be observed directly with the human eye.

The first line of the Balmer series, known as Hα, corresponds to electron transitions from n=3 to n=2. This visible line contributes to the rainbow of spectral colors seen when examining objects like stars or planets through a spectroscope.

Notable facts about the Balmer series:
  • Involves transitions down to the second principal energy level (n=2).
  • Visible in the optical range, unlike the Lyman series.
  • Frequently observed in astronomical and laboratory spectroscopy.
This series offers vital insights into physical conditions in space, such as temperature and density of interstellar clouds.
Rydberg Formula
The Rydberg formula is a pioneering tool in atomic physics used to calculate the wavelengths (and thus frequencies and energies) of light resulting from electron transitions in hydrogen. This formula is particularly beneficial for identifying the spectral lines of any series, such as the Lyman, Balmer, and others.

The general form of the Rydberg formula is given by:\[ u = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\]where:
  • \( u \) is the frequency of the emitted or absorbed light.
  • \( R_H \) represents the Rydberg constant, a fundamental value for hydrogen.
  • \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and higher energy levels involved.
This formula shows how frequency relates directly to the energy shifts between different levels within the hydrogen atom. Understanding and applying the Rydberg formula is essential for distinguishing different spectral lines and verifying phenomena like the Ritz combination principle, which links different lines by displaying their additive properties.

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Most popular questions from this chapter

Light of wavelength \(102.6 \mathrm{~nm}\) is emitted by a hydrogen atom. What are the (a) higher quantum number and (b) lower quantum number of the transition producing this emission? (c) What is the name of the series that includes the transition?

The radial probability density for the ground state of the hydrogen atom is a maximum when \(r=a\), where \(a\) is the Bohr radius. Show that the average value of \(r\), defined as $$ r_{\text {avg }}=\int P(r) r d r $$ has the value \(1.5 a\). In this expression for \(r_{\text {avg }}\), each value of \(P(r)\) is weighted with the value of \(r\) at which it occurs. Note that the average value of \(r\) is greater than the value of \(r\) for which \(P(r)\) is a maximum.

An old model of a hydrogen atom has the charge \(+e\) of the proton uniformly distributed over a sphere of radius \(a_{0}\), with the electron of charge \(-e\) and mass \(m\) at its center. (a) What would then be the force on the electron if it were displaced from the center by a distance \(r \leq a_{0} ?\) (b) What would be the angular frequency of oscillation of the electron about the center of the atom once the electron was released?

In a simple model of a hydrogen atom, the single electron orbits the single proton (the nucleus) in a circular path. Calculate (a) the electric potential set up by the proton at the orbital radius of \(52.9 \mathrm{pm},(\mathrm{b})\) the electric potential energy of the atom, and (c) the kinetic energy of the electron. (d) How much energy is required to ionize the atom (that is, to remove the electron to an infinite distance with no kinetic energy)? Give the energies in electron-volts.

What must be the width of a one-dimensional infinite potential well if an electron trapped in it in the \(n=3\) state is to have an energy of \(4.7 \mathrm{eV} ?\)

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