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

For your work in a mass spectrometry lab, you are investigating the absorption spectrum of one-electron ions. To maintain the atoms in an ionized state, you hold them at low density in an ion trap, a device that uses a configuration of electric fields to confine ions. The majority of the ions are in their ground state, so that is the initial state for the absorption transitions that you observe. (a) If the longest wavelength that you observe in the absorption spectrum is 13.56 nm, what is the atomic number Z for the ions? (b) What is the next shorter wavelength that the ions will absorb? (c) When one of the ions absorbs a photon of wavelength 6.78 nm, a free electron is produced. What is the kinetic energy (in electron volts) of the electron?

Short Answer

Expert verified
(a) Calculate \( Z \) using the Rydberg formula. (b) Use the next transition formula for \( Z^2 \) found. (c) Photon energy minus ionization energy gives electron kinetic energy.

Step by step solution

01

Understand the Ion Configuration

Since we are dealing with one-electron ions, we should note that the spectrum is similar to that of the hydrogen atom but with adjustments depending on the atomic number \( Z \). The longest wavelength corresponds to the transition from the first excited state (\( n=2 \)) to the ground state (\( n=1 \)).
02

Calculate Atomic Number (\( Z \))

Using the Rydberg formula \( \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) for the longest wavelength at \( \lambda = 13.56 \text{ nm} = 13.56 \times 10^{-9} \text{ m} \), \( n_1 = 1 \), \( n_2 = 2 \), and the Rydberg constant \( R = 1.097 \times 10^7 \text{ m}^{-1} \), solve for \( Z \):\[ Z^2 = \frac{1}{R \lambda} \frac{4}{3} \]Calculate \( Z \).
03

Determine the Next Shorter Wavelength

The next shorter wavelength will correspond to a transition from \( n_2=3 \) to \( n_1=1 \). Use \[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \] to find this wavelength.\[ Z^2 \] is already calculated from the previous step.
04

Convert Wavelength to Energy for Ionization

For (c), calculate the energy of a photon with wavelength \( 6.78 \text{ nm} \) using \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \text{ Js} \) and \( c = 3 \times 10^8 \text{ m/s} \). Convert this energy to electron volts by dividing by \( eV, 1.6 \times 10^{-19} \text{ J/eV} \).
05

Calculate Kinetic Energy of the Electron

Subtract the ionization energy corresponding to the ground state to free electron (i.e., full ionization with \( n_1=1, n_2=\infty \)) from the previously calculated photon energy to find the kinetic energy of the electron.

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.

Understanding Mass Spectrometry
Mass spectrometry is a powerful tool used to analyze ions based on their mass-to-charge ratio. It can provide detailed information about the composition and abundance of ions. In this exercise, we're looking at one-electron ions confined in an ion trap to maintain them in an ionized state.
  • Ion Trap: An ion trap uses electric fields to keep ions stable and in place, which is crucial for precise measurements in mass spectrometry.
  • Mass-to-Charge Ratio: The primary information obtained from mass spectrometry, critical for distinguishing between different ions.
By understanding mass spectrometry in this context, we gain insight into how the ions are analyzed as they transition between energy states.
Exploring Absorption Spectrum
The absorption spectrum of an element shows the specific wavelengths of light absorbed by the electrons of its atoms, which corresponds to transitions between energy levels. In the case of one-electron ions, these spectra resemble the hydrogen spectrum but are adjusted based on the atomic number.
  • Ground State: Most of the ions are in this lowest energy state initially, requiring energy absorption to transition to a higher state.
  • Longest Wavelength: Represents the smallest energy difference, typically seen in transitions from the first excited state to the ground state.
This spectrum aids in calculating the atomic number and understanding electron transitions for the ions in question.
Explaining the Rydberg Formula
The Rydberg formula helps calculate the wavelengths of spectral lines of one-electron ions, adjusting for the atomic number. It’s expressed as:\[ \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
  • Rydberg Constant (\( R \)): A fundamental constant in physics, typically \( 1.097 \times 10^7 \text{ m}^{-1} \).
  • Quantum Numbers: \( n_1 \) and \( n_2 \) represent the initial and final energy levels of the electron transition, respectively.
Using this formula, we can calculate important characteristics like the atomic number or other photon interactions with the ions.
Performing Atomic Number Calculation
Calculating the atomic number \( Z \) of an ion involves analyzing the energy transitions within its electron configuration. For one-electron ions, knowing the wavelength of absorbed light allows for \( Z \) determination using the Rydberg formula.
  • Variables Involved: Knowing the wavelengths and applying the appropriate energy levels (\( n_1, n_2 \)).
  • Solving for \( Z \): By rearranging the Rydberg formula to isolate \( Z \), we can calculate its value accurately using the known line spectrum data.
This precise calculation is foundational to understanding the ion's energy transitions and its behavior in such an experiment.

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

Light from an ideal spherical blackbody 15.0 cm in diameter is analyzed by using a diffraction grating that has 3850 lines/cm. When you shine this light through the grating, you observe that the peak-intensity wavelength forms a first-order bright fringe at \(\pm\)14.4\(^\circ\) from the central bright fringe. (a) What is the temperature of the blackbody? (b) How long will it take this sphere to radiate 12.0 MJ of energy at constant temperature?

(a) For one-electron ions with nuclear charge Z, what is the speed of the electron in a Bohr-model orbit labeled with \(n\)? Give your answer in terms of \(v_1\), the orbital speed for the \(n\) = 1 Bohr orbit in hydrogen. (b) What is the largest value of Z for which the \(n\) = 1 orbital speed is less than 10\(\%\) of the speed of light in vacuum?

A 4.78-MeV alpha particle from a \(^{226}\)Ra decay makes a head-on collision with a uranium nucleus. A uranium nucleus has 92 protons. (a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uranium nucleus remains at rest and that the distance of closest approach is much greater than the radius of the uranium nucleus. (b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?

A sample of hydrogen atoms is irradiated with light with wavelength 85.5 nm, and electrons are observed leaving the gas. (a) If each hydrogen atom were initially in its ground level, what would be the maximum kinetic energy in electron volts of these photoelectrons? (b) A few electrons are detected with energies as much as 10.2 eV greater than the maximum kinetic energy calculated in part (a). How can this be?

An alpha particle (\(m = 6.64 \times 10^{-27} kg\)) emitted in the radioactive decay of uranium-238 has an energy of 4.20 MeV. What is its de Broglie wavelength?

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