Concept explainers
In the Stern-Gerlach experiment, silver atoms were used. This was a good choice, as it turned out. Using the electron configuration of silver atoms, explain why silver was a good candidate for being able to observe the intrinsic
Interpretation:
The reason as to why silver was a good candidate for being able to observe the intrinsic angular momentum of the electron in the Stern-Gerlach experiment is to be stated.
Concept introduction:
Stern and Gerlach performed an experiment in which a beam of silver atoms was passed through a magnetic field. They observed that the beam of silver atoms was splitted into two different beams. The result of the experiment concluded that every electron has an intrinsic angular momentum that they termed it as spin angular momentum.
Answer to Problem 12.1E
Silver was a good candidate for being able to observe the intrinsic angular momentum of the electron in the Stern-Gerlach experiment as it has only one unpaired electron in its
Explanation of Solution
The electron configuration of
The presence of an unpaired electron in silver makes its interaction possible with the external magnetic field. The interaction between spin of this unpaired electron and external magnetic field resulted in the discovery of intrinsic angular momentum of the electron.
Silver was a good candidate for being able to observe the intrinsic angular momentum of the electron in the Stern-Gerlach experiment as it has only one unpaired electron in its
Want to see more full solutions like this?
Chapter 12 Solutions
Physical Chemistry
- What experimental evidence supports the quantum theory of light? Explain the wave-particle duality of all matter .. For what size particles must one consider both the wave and the particle properties?arrow_forwardInvestigating Energy Levels Consider the hypothetical atom X that has one electron like the H atom but has different energy levels. The energies of an electron in an X atom are described by the equation E=RHn3 where RH is the same as for hydrogen (2.179 1018 J). Answer the following questions, without calculating energy values. a How would the ground-state energy levels of X and H compare? b Would the energy of an electron in the n = 2 level of H be higher or lower than that of an electron in the n = 2 level of X? Explain your answer. c How do the spacings of the energy levels of X and H compare? d Which would involve the emission of a higher frequency of light, the transition of an electron in an H atom from the n = 5 to the n = 3 level or a similar transition in an X atom? e Which atom, X or H, would require more energy to completely remove its electron? f A photon corresponding to a particular frequency of blue light produces a transition from the n = 2 to the n = 5 level of a hydrogen atom. Could this photon produce the same transition (n = 12 to n = 5) in an atom of X? Explain.arrow_forwardWhy is the electron in a Bohr hydrogen atom bound less tightly when it has a quantum number of 3 than when it has a quantum number of 1?arrow_forward
- How is the Bohr theory of the hydrogen atom inconsistent with the uncertainty principle? In fact, it was this inconsistency, along with the theorys limited application to non-hydrogen-like systems, that limited Bohrs theory.arrow_forward• identify an orbital (as 1s, 3p, etc.) from its quantum numbers, or vice versa.arrow_forwardWhich of the following sets of quantum numbers correctly represents a 4p orbital? (a) n = 4, = 0, m = 1 (b) n = 4, = 1, m = 0 (c) n = 4, = 2, m = 1 (d) n = 4, = 1, m =2arrow_forward
- Suppose that the spin quantum number did not exist, and therefore only one electron could occupy each orbital of a many-electron atom. Give the atomic numbers of the first three noble-gas atoms in this case.arrow_forward6.92 The photoelectric effect can he used to measure the value of Planck's constant. Suppose that a photoelectric effect experiment was carried out using light with v=7.501014s1 and ejected electrons were detected with a kinetic energy of 2.501011 J. The experiment was then repeated using light with v=1.001015s1 and the same metal target, and electrons were ejected with kinetic energy of 5.001011 J. Use these data to find a value for Planck’s constant. HINTS: These data are fictional and will give a result that is quite different from the real value of Planck's constant. Be sure that you do not use the real value of Planck's constant in any calculations here. It may help to start by thinking about how you would calculate the metal's binding energy if you already knew Planck's constant.arrow_forwardAs the weapons officer aboard the Srarship Chemistry, it is your duty to configure a photon torpedo to remove an electron from the outer hull of an enemy vessel. You know that the work function (the binding energy of the electron) of the hull of the enemy ship is 7.52 1019 J. a. What wavelength does your photon torpedo need to be to eject an electron? b. You find an extra photon torpedo with a wavelength of 259 nm and fire it at the enemy vessel. Does this photon torpedo do any damage to the ship (does it eject an electron)? c. If the hull of the enemy vessel is made of the element with an electron configura tion of [Ar]4s13d10, what metal is this?arrow_forward
- 6.32 What are the mathematical origins of quantum numbers?arrow_forward6.96 When a helium atom absorbs light at 58.44 nm, an electron is promoted from the 1s orbital to a 2p orbital. Given that the ionization energy of (ground state) helium is 2372 kJ/ mol, find the longest wavelength of light that could eject an electron from the excited state helium atom.arrow_forward6.93 A mercury atom is initially in its lowest possible (or ground state) energy level. The atom absorbs a photon with a wavelength of 185 nm and then emits a photon with a frequency of 4.9241014HZ . At the end of this series of transitions, the atom will still be in an energy level above the ground state. Draw an energy-level diagram for this process and find the energy of this resulting excited state, assuming that we assign a value of E = 0 to the ground state. (This choice of E = 0 is not the usual convention, but it will simplify the calculations you need to do here.)arrow_forward
- Chemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage LearningIntroductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
- Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning