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Use the de Broglie relationship to determine the wavelengths of the following objects: (a) an \(85-\mathrm{kg}\) person skiing at \(50 \mathrm{~km} / \mathrm{hr},\) (b) a 10.0 -g bullet fired at \(250 \mathrm{~m} / \mathrm{s},(\mathrm{c})\) a lithium atom moving at \(2.5 \times 10^{5} \mathrm{~m} / \mathrm{s},(\mathrm{d})\) an ozone \(\left(\mathrm{O}_{3}\right)\) molecule in the upper atmosphere moving at \(550 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The de Broglie wavelengths of the given objects are: (a) an 85-kg person skiing at 50 km/hr has a wavelength of \(5.48 \times 10^{-38}\: \text{m}\), (b) a 10.0-g bullet fired at 250 m/s has a wavelength of \(2.65 \times 10^{-34}\: \text{m}\), (c) a lithium atom moving at \(2.5 \times 10^{5}\: \text{m/s}\) has a wavelength of \(2.30 \times 10^{-10}\: \text{m}\), and (d) an ozone molecule (O3) in the upper atmosphere moving at 550 m/s has a wavelength of \(1.47 \times 10^{-14}\: \text{m}\).

Step by step solution

01

Setup the De Broglie equation and constants values

First, let's write the De Broglie equation and list the constant values we will be using. \( \lambda = \frac{h}{mv} \) Planck's constant, \( h = 6.63 \times 10^{-34} \: \text{Js} \) (a) An 85 kg person skiing at 50 km/hr (b) A 10.0 g bullet fired at 250 m/s (c) A lithium atom moving at \(2.5 \times 10^{5}\: \text{m/s} \) (d) An ozone molecule (O3) in the upper atmosphere moving at 550 m/s
02

Solve for (a) An 85 kg person skiing at 50 km/hr

First, we need to convert the velocity from km/hr to m/s: \( 50\: \text{km/hr} = \frac{50 \times 1000}{3600}\: \text{m/s} \) \( v = 13.89\: \text{m/s} \) Now, we can plug this value and given mass into the de Broglie equation: \( \lambda = \frac{6.63 \times 10^{-34}\: \text{Js}}{85\: \text{kg} × 13.89\: \text{m/s}} \) \( \lambda \approx 5.48 \times 10^{-38}\: \text{m} \)
03

Solve for (b) A 10.0 g bullet fired at 250 m/s

First, we need to convert the mass from grams to kilograms: 10 g = 0.01 kg Now, use the de Broglie equation: \( \lambda = \frac{6.63 \times 10^{-34}\: \text{Js}}{0.01\: \text{kg} × 250\: \text{m/s}} \) \( \lambda \approx 2.65 \times 10^{-34}\: \text{m} \)
04

Solve for (c) A lithium atom moving at \(2.5 \times 10^{5}\: \text{m/s} \)

The mass of a lithium atom can be calculated using the atomic mass, which is 6.93 amu. To convert to kg, multiply by the conversion factor: \( m = 6.93 \: \text{amu} × \frac{1.66 \times 10^{-27}\: \text{kg}}{1\: \text{amu}} \) \( m = 1.15 \times 10^{-26}\: \text{kg} \) Now, use the de Broglie equation: \( \lambda = \frac{6.63 \times 10^{-34}\: \text{Js}}{1.15 \times 10^{-26}\: \text{kg} × 2.5 \times 10^{5}\: \text{m/s}} \) \( \lambda \approx 2.30 \times 10^{-10}\: \text{m} \)
05

Solve for (d) An ozone molecule (O3) in the upper atmosphere moving at 550 m/s

The mass of an O3 molecule can be determined by finding the total mass of 3 oxygen atoms. The atomic mass of oxygen is 16.00 amu, and the O3 molecule has a mass of: \( m_{O_3} = 3 \times 16.00\: \text{amu} = 48.00\: \text{amu} \) Now, convert this mass from amu to kg: \( m = 48.00 \: \text{amu} × \frac{1.66 \times 10^{-27}\: \text{kg}}{1\: \text{amu}} \) \( m = 7.97 \times 10^{-26}\: \text{kg} \) Now, use the de Broglie equation: \( \lambda = \frac{6.63 \times 10^{-34}\: \text{Js}}{7.97 \times 10^{-26}\: \text{kg} × 550\: \text{m/s}} \) \( \lambda \approx 1.47 \times 10^{-14}\: \text{m} \)

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Key Concepts

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

Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that describes the physical properties of nature at the smallest scales, such as atoms and subatomic particles. It represents a departure from classical mechanics and introduces new principles, such as quantization of energy, wave-particle duality, and the uncertainty principle.

At the core of quantum mechanics is the idea that particles can exhibit both wave-like and particle-like properties. This theory not only explains the behavior of the micro-world but also assists in understanding macroscopic phenomena. One of the essential tools within quantum mechanics is the de Broglie hypothesis, which associates a wavelength with any moving particle—a concept that was demonstrated in the exercise with objects that have mass and are in motion, like a person skiing or a bullet being fired.

Understanding quantum mechanics is crucial for comprehending a myriad of modern technologies, including semiconductors, lasers, and even medical imaging devices such as MRI. The field is expansive and often counterintuitive, challenging our classical viewpoints on how the world operates.
Wave-Particle Duality
Wave-particle duality is the concept that all particles exhibit both wave and particle properties. This dual characteristic is a cornerstone of quantum mechanics and was first put forth by French physicist Louis de Broglie in 1924.

In the exercise, this duality is illustrated through the de Broglie wavelength calculation for different particles and objects. The key equation, \( \lambda = \frac{h}{mv} \), succinctly encapsulates this duality by relating the momentum of a particle (\(mv\)) to a characteristic wave property, the wavelength (\(\lambda\)). For example, the calculations showed that massive objects, like a skier or a bullet, have exceedingly short wavelengths that are not detectable in everyday life, highlighting the fact that wave-like behavior becomes significant only at the atomic or subatomic levels.

Wave-particle duality requires us to abandon the classic distinction between waves and particles; instead, we must adopt a more holistic approach to understanding the nature and behavior of quantum objects. This duality is experimentally observed in phenomena such as electron diffraction and the famous double-slit experiment.
Planck's Constant
Planck's constant (denoted as \(h\)) is a fundamental constant in the realm of quantum mechanics with the value of \(6.63 \times 10^{-34} \text{Js}\). It is named after the physicist Max Planck who discovered it in 1900 while solving a problem in black-body radiation.

Planck's constant is essential in the quantization of physical quantities such as energy, momentum, and angular momentum. It serves as the bridge between the macroscopic and microscopic worlds by setting the scale at which quantum effects become significant. For instance, in the de Broglie equation mentioned in the original exercise, Planck's constant connects a moving object's mass and velocity to its wavelength, providing evidence of quantum behavior.

Planck's constant also appears in the famous equation for the energy of a photon (\(E = hf\)), where \(f\) is the frequency of the photon. This relation has profound implications for understanding the interactions between matter and electromagnetic radiation, the structure of atoms, the nature of chemical bonds, and forms the basis for technologies such as atomic clocks and quantum computing. It is a reminder of the discrete, or 'quantized', nature of the universe at a fundamental level.

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Most popular questions from this chapter

(a) What is the frequency of radiation that has a wavelength of \(10 \mu \mathrm{m},\) about the size of a bacterium? (b) What is the wavelength of radiation that has a frequency of \(5.50 \times 10^{14} \mathrm{~s}^{-1}\) ? (c) Would the radiations in part (a) or part (b) be visible to the human eye? (d) What distance does electromagnetic radiation travel in \(50.0 \mu \mathrm{s} ?\)

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, \(\mathrm{ZrO}_{2}\), is reacted with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and CO in the ratio 1: 2 . Write a balanced chemical equation for the reaction. Starting with a 55.4-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 \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\)

For a given value of the principal quantum number, \(n\), how do the energies of the \(s, p, d,\) and \(f\) subshells vary for (a) hydrogen, (b) a many-electron atom?

(a) Why does the Bohr model of the hydrogen atom violate the uncertainty principle? (b) In what way is the description of the electron using a wave function consistent with de Broglie's hypothesis? (c) What is meant by the term probability density? Given the wave function, how do we find the probability density at a certain point in space?

State where in the periodic table these elements appear: (a) elements with the valence-shell electron configuration \(n s^{2} n p^{5}\) (b) elements that have three unpaired \(p\) electrons (c) an element whose valence electrons are \(4 s^{2} 4 p^{1}\) (d) the \(d\) -block elements

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