(b) = 0.18 cm In this case the wavelength is significant. While the De Broglie equation applies to all systems, the wave properties become observable only for microscopic objects. 541

An application of De Broglie’s idea 542

An application of De Broglie’s idea A key idea from optics: You can “see” an object about ½ the size of the wavelength of light used. 543

An application of De Broglie’s idea A key idea from optics: You can “see” an object about ½ the size of the wavelength of light used. So for visible light at 400 nm, you could see an object with a size of around 200 nm or 2x10 -5 cm. 544

An application of De Broglie’s idea A key idea from optics: You can “see” an object about ½ the size of the wavelength of light used. So for visible light at 400 nm, you could see an object with a size of around 200 nm or 2x10 -5 cm. To “see” something smaller, could use X-rays – but its hard to focus X-rays. 545

However, could work with electrons in place of light. 546

However, could work with electrons in place of light. Relatively easy to accelerate electrons to high velocity, so from the formula above, the wavelength would be small. 547

However, could work with electrons in place of light. Relatively easy to accelerate electrons to high velocity, so from the formula above, the wavelength would be small. Can now see individual atoms. 548

The Heisenberg Uncertainty Principle 549

The Heisenberg Uncertainty Principle In 1927 Heisenberg proposed a principle of great importance in the philosophical groundwork of quantum theory. 550

The Heisenberg Uncertainty Principle In 1927 Heisenberg proposed a principle of great importance in the philosophical groundwork of quantum theory. Heisenberg’s principle is primarily concerned with measurements on the atomic scale. 551

In classical physics it was generally believed that the accuracy with which any quantity could be determined depended upon the instrument used to do the measurements. 552

In classical physics it was generally believed that the accuracy with which any quantity could be determined depended upon the instrument used to do the measurements. Thought experiment: Suppose we wish to monitor the position and velocity of a tennis ball in flight. Can do this with different instruments. Important consideration – must be able to “see” the tennis ball – simply shine light on the ball. 553

Because photons (in the visible part of the electromagnetic spectrum) have small momentum, the impact they have on the tennis ball is negligibly small. 554

Because photons (in the visible part of the electromagnetic spectrum) have small momentum, the impact they have on the tennis ball is negligibly small. Consider the corresponding experiment on an electron in flight. Must be able to “see” the electron. This requires light with a very small wavelength. Recall the two equations: 555

Because photons (in the visible part of the electromagnetic spectrum) have small momentum, the impact they have on the tennis ball is negligibly small. Consider the corresponding experiment on an electron in flight. Must be able to “see” the electron. This requires light with a very small wavelength. Recall the two equations: E = h and hence. 556

Because photons (in the visible part of the electromagnetic spectrum) have small momentum, the impact they have on the tennis ball is negligibly small. Consider the corresponding experiment on an electron in flight. Must be able to “see” the electron. This requires light with a very small wavelength. Recall the two equations: E = h and hence. Therefore the light has high energy if the wavelength is very small. 557

The impact of high energy photons on an electron can appreciably alter the flight path of the electron. That is, the electron’s velocity and its momentum are altered. 558

The impact of high energy photons on an electron can appreciably alter the flight path of the electron. That is, the electron’s velocity and its momentum are altered. Thus at the very instant we determine the position of the electron – we lose information on its momentum. 559

The impact of high energy photons on an electron can appreciably alter the flight path of the electron. That is, the electron’s velocity and its momentum are altered. Thus at the very instant we determine the position of the electron – we lose information on its momentum. To reduce the impact we could choose light of a longer wavelength (i.e. lower energy), but then we would be unable to pinpoint the position of the electron. 560

This kind of reasoning led Heisenberg to the Heisenberg Uncertainty Principle: 561

This kind of reasoning led Heisenberg to the Heisenberg Uncertainty Principle: It is impossible to determine to arbitrary accuracy both the position and momentum of a particle at the same time. 562

This kind of reasoning led Heisenberg to the Heisenberg Uncertainty Principle: It is impossible to determine to arbitrary accuracy both the position and momentum of a particle at the same time. Let x represent the position and p the momentum of a particle. Let and denote the uncertainties in the measurements (i.e. the measurement errors), then 563

Heisenberg showed that 564

Heisenberg showed that Of course it always possible to do a sloppy experiment, then the product of the two uncertainties could be a lot larger than. 565

Heisenberg showed that Of course it always possible to do a sloppy experiment, then the product of the two uncertainties could be a lot larger than. Like the De Broglie relation, the Heisenberg Uncertainty Principle applies to both macroscopic and microscopic objects. 566

The Heisenberg Uncertainty Principle sets an inherent upper limit for measuring sub- microscopic objects. 567

The Heisenberg Uncertainty Principle sets an inherent upper limit for measuring sub- microscopic objects. Consequently, it is impossible to pin down the exact nature of the electron. 568

The Heisenberg Uncertainty Principle sets an inherent upper limit for measuring sub- microscopic objects. Consequently, it is impossible to pin down the exact nature of the electron. Whether an electron should be thought of as a particle, a wave, or a particle-wave – depends on how we take a measurement. 569

When we speak of size, mass, and charge of an electron – we are thinking about its particle properties. 570

When we speak of size, mass, and charge of an electron – we are thinking about its particle properties. When we speak of wavelength of an electron – we are talking about the electron as a wave. 571

When we speak of size, mass, and charge of an electron – we are thinking about its particle properties. When we speak of wavelength of an electron – we are talking about the electron as a wave. There are some additional properties of the electron that we will encounter that will tie in directly with the wave-like nature of the electron. 572

Problem Example, Heisenberg Uncertainty Principle: The uncertainty in measuring the velocity of an electron is 1.0 x cm s -1. Calculate the corresponding uncertainty in determining its position. 573

Problem Example, Heisenberg Uncertainty Principle: The uncertainty in measuring the velocity of an electron is 1.0 x cm s -1. Calculate the corresponding uncertainty in determining its position. Now p = m v 574

Problem Example, Heisenberg Uncertainty Principle: The uncertainty in measuring the velocity of an electron is 1.0 x cm s -1. Calculate the corresponding uncertainty in determining its position. Now p = m v so = m 575

Problem Example, Heisenberg Uncertainty Principle: The uncertainty in measuring the velocity of an electron is 1.0 x cm s -1. Calculate the corresponding uncertainty in determining its position. Now p = m v so = m = (9. 11 x g)(1.0 x cm s -1 ) = 9.1 x g cm s

From, for the purpose of calculation, we can treat the as an = sign. Hence, 577

From, for the purpose of calculation, we can treat the as an = sign. Hence, 578

From, for the purpose of calculation, we can treat the as an = sign. Hence, = 5.8 x 10 3 m (or 3.6 miles!) 579

Exercise: Try a similar calculation on your instructor (or yourself)! Assume a mass of around 80 kg (or your mass) and an uncertainty in velocity of around 0.5 cm s -1. Is the uncertainty in the instructor’s position (your position) so uncertain that he (you) is (are) not in the room? 580

Quantum Mechanics The spectacular initial success of Bohr’s theory was followed by a series of disappointments. 581

Quantum Mechanics The spectacular initial success of Bohr’s theory was followed by a series of disappointments. Bohr (nor anyone else) could account for the emission spectra of atoms more complex than the H atom. 582

It was realized eventually that a new theory was required. A new equation beyond Newton’s laws of motion was needed to describe the microscopic world. 583

It was realized eventually that a new theory was required. A new equation beyond Newton’s laws of motion was needed to describe the microscopic world. The key figure to make the great advance was Schrödinger (1926). 584

It was realized eventually that a new theory was required. A new equation beyond Newton’s laws of motion was needed to describe the microscopic world. The key figure to make the great advance was Schrödinger (1926). Schrödinger formulated a new equation to describe the motion of microscopic bodies. 585

Schrödinger formulated an equation that incorporated the particle properties in terms of the mass (m) and charge (e) and the wave properties in terms of a function – which is called the wave function. The symbol employed for the wave function is (psi). 586

Schrödinger formulated an equation that incorporated the particle properties in terms of the mass (m) and charge (e) and the wave properties in terms of a function – which is called the wave function. The symbol employed for the wave function is (psi). There is no simple physical meaning for the function. However the function has the meaning that it gives the probability of finding the particle at a particular point in space. 587

This probability thinking is a radical break with classical thinking. 588

The Hydrogen atom and the Schr ö dinger equation 589

The Hydrogen atom and the Schr ö dinger equation One of the first problems investigated by Schrödinger was the energy levels for the H atom. 590

The Hydrogen atom and the Schr ö dinger equation One of the first problems investigated by Schrödinger was the energy levels for the H atom. Schrödinger obtained the result: 591

592

where: m = the mass of the electron 593

where: m = the mass of the electron e = charge on the electron 594

where: m = the mass of the electron e = charge on the electron h = Planck’s constant 595

where: m = the mass of the electron e = charge on the electron h = Planck’s constant n = a positive integer value ( n = 1, 2, 3, …) called the principal quantum number 596

where: m = the mass of the electron e = charge on the electron h = Planck’s constant n = a positive integer value ( n = 1, 2, 3, …) called the principal quantum number = vacuum permittivity 597

where: m = the mass of the electron e = charge on the electron h = Planck’s constant n = a positive integer value ( n = 1, 2, 3, …) called the principal quantum number = vacuum permittivity E n = the energy of the different states 598

This formula shows that the energies of an electron in the hydrogen atom are quantized. 599

This formula shows that the energies of an electron in the hydrogen atom are quantized. The above formula will be abbreviated as k E n = – n 2 600

It should be no surprise that the Schrödinger result exactly matches the Bohr result. Recall that Bohr was able to explain the spectrum of the H atom, exactly matching the best experimental data available at the time. 601

Explanation of the spectrum of the H atom 602

Explanation of the spectrum of the H atom Energy Level Diagram for the H atom 603

Explanation of the spectrum of the H atom Energy Level Diagram for the H atom Consider two energy levels for the H atom, the upper one characterized by the quantum number n upper (abbreviated n u ) and a lower level characterized by the quantum number n lower (abbreviated n l ). 604

Two situations: (1) absorption and (2) emission. n u n l absorption (photon absorbed) 605

n u n l emission (photon emitted) 606

In both the absorption and emission processes Recall so that 607

In both the absorption and emission processes Recall so that k From the formula E n = - we can n 2 write : 608

and 609

and Since we have 610

and Since we have 611

and Since we have 612

and Since we have 613

That is 614

That is Set 615

That is Set So that 616

That is Set So that This is called the Rydberg formula and R H is called the Rydberg constant. 617

Problem Example: What is the wavelength and frequency of photons emitted during a transition from n u = 5 state to the n l =2 state in the H atom? Also calculate the energy difference between these two states. 618

Problem Example: What is the wavelength and frequency of photons emitted during a transition from n u = 5 state to the n l =2 state in the H atom? Also calculate the energy difference between these two states. R H = cm -1 and k = x J 619

Problem Example: What is the wavelength and frequency of photons emitted during a transition from n u = 5 state to the n l =2 state in the H atom? Also calculate the energy difference between these two states. R H = cm -1 and k = x J Approach 1 620

Problem Example: What is the wavelength and frequency of photons emitted during a transition from n u = 5 state to the n l =2 state in the H atom? Also calculate the energy difference between these two states. R H = cm -1 and k = x J Approach 1 621

Problem Example: What is the wavelength and frequency of photons emitted during a transition from n u = 5 state to the n l =2 state in the H atom? Also calculate the energy difference between these two states. R H = cm -1 and k = x J Approach 1 622

= cm

= cm -1 Hence = x10 -5 cm = nm 624

= cm -1 Hence = x10 -5 cm = nm Now use 625

= x J 626

= x J Approach 2 Use k E n = - n 2 627

So that k k E 5 = - E 2 =

So that k k E 5 = - E 2 = = x J = x J 629

So that k k E 5 = - E 2 = = x J = x J Therefore = E 5 - E 2 = x J 630