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Chapter 6: Problem 36

Classify each of the following statements as either true or false: (a) A hydrogen atom in the \(n=3\) state can emit light at only two specific wavelengths, \((\mathbf{b})\) a hydrogen atom in the \(n=2\) state is at a lower energy than one in the \(n=1\) state, and (c) the energy of an emitted photon equals the energy difference of the two states involved in the emission.

Short Answer

Expert verified
Based on the analysis of given statements: (a) This statement is \(\mathbf{True}\), because an electron in the n=3 state can emit light at only two specific wavelengths. (b) This statement is \(\mathbf{False}\), as a hydrogen atom in the n=1 state is at lower energy than one in the n=2 state. (c) This statement is \(\mathbf{True}\), as the energy of an emitted photon equals the energy difference of the two states involved in the emission.

Step by step solution

01

Statement (a) Classification

We are asked to determine if a hydrogen atom in the n=3 can emit light at only two specific wavelengths. To analyze this, we have to recall that when an electron transitions between states, the energy differences correspond to specific photon energies, which in turn correspond to specific wavelengths. When an electron in the n=3 state transitions to a lower-energy state, the possible lower-energy states are n=2 and n=1. So, we have: - Transition from n=3 to n=2 - Transition from n=3 to n=1 Hence, an electron in the n=3 state can emit light at only two specific wavelengths. This statement is therefore \(\mathbf{True}\).
02

Statement (b) Classification

We are asked to determine if a hydrogen atom in the n=2 state is at a lower energy than one in the n=1 state. To analyze this, we have to recall the formula for the energy levels of hydrogen state, which is given by: \[ E_n = - \frac{13.6 eV}{n^2} \] Where n is the state, and E is the energy level. For n=1, we have: \(E_1 = - \frac{13.6 eV}{1^2} = -13.6 eV \) For n=2, we have: \(E_2 = - \frac{13.6 eV}{2^2} = -\frac{13.6 eV}{4} = -3.4 eV \) Since -13.6 eV is less than -3.4 eV, a hydrogen atom in the n=1 state is at lower energy than one in the n=2 state. This statement is therefore \(\mathbf{False}\).
03

Statement (c) Classification

We are asked to determine if the energy of an emitted photon equals the energy difference of the two states involved in the emission. To analyze this, we have to recall that when an electron transitions from a higher energy state to a lower energy state, it emits a photon with the exact energy difference between the two states. The energy difference is given by: \( \Delta E = E_\text{final} - E_\text{initial} \) Since the energy of the emitted photon equals the energy difference between the two states, this statement is therefore \(\mathbf{True}\).

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Most popular questions from this chapter

In the experiment shown schematically below, a beam of neutral atoms is passed through a magnetic field. Atoms that have unpaired electrons are deflected in different directions in the magnetic field depending on the value of the electron spin quantum number. In the experiment illustrated, we envision that a beam of hydrogen atoms splits into two beams. (a) What is the significance of the observation that the single beam splits into two beams? (b) What do you think would happen if the strength of the magnet were increased? (c) What do you think would happen if the beam of hydrogen atoms were replaced with a beam of helium atoms? Why? (d) The relevant experiment was first performed by Otto Stern and Walter Gerlach in \(1921 .\) They used a beam of \(\mathrm{Ag}\) atoms in the experiment. By considering the electron configuration of a silver atom, explain why the single beam splits into two beams.

Identify the group of elements that corresponds to each of the following generalized electron configurations and indicate the number of unpaired electrons for each: (a) [noble gas]ns \(^{2} n p^{5}\) (b) \([\) noble gas \(] n s^{2}(n-1) d^{2}\) (c) \([\) noble gas \(] n s^{2}(n-1) d^{10} n p^{1}\) (d) \([\) noble \(\operatorname{gas}] n s^{2}(n-2) f^{6}\)

In the television series Star Trek, the transporter beam is a device used to "beam down" people from the Starship Enterprise to another location, such as the surface of a planet. The writers of the show put a "Heisenberg compensator" into the transporter beam mechanism. Explain why such a compensator (which is entirely fictional) would be necessary to get around Heisenberg's uncertainty principle.

(a) What are the similarities of and differences between the \(1 s\) and \(2 s\) orbitals of the hydrogen atom? (b) In what sense does a \(2 p\) orbital have directional character? Compare the "directional" characteristics of the \(p_{x}\) and \(d_{x^{2}-y^{2}}\) orbitals. (That is, in what direction or region of space is the electron density concentrated?) (c) What can you say about the average distance from the nucleus of an electron in a \(2 s\) orbital as compared with a 3 s orbital? ( \(\mathbf{d}\) ) For the hydrogen atom, list the following orbitals in order of increasing energy (that is, most stable ones first): \(4 f, 6 s, 3 d, 1 s, 2 p\)

The discovery of hafnium, element number \(72,\) provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements \(58-71\) ) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), reacts with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2 .\) Write a balanced chemical equation for the reaction. Starting with a \(55.4-\mathrm{g}\) sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and Hf form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).
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