Physics 1C Lecture 29A. Finish off Ch. 28 Start Ch. 29

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1 Physics 1C Lecture 29A Finish off Ch. 28 Start Ch. 29

2 Particle in a Box Let s consider a particle confined to a one-dimensional region in space. Following the quantum mechanics approach, we need to find an appropriate wave function to describe the motion of the particle.! Because of the walls, the probability of finding the particle outside the box is zero.! This means that ψ(x)=0 for x 0 and for x L.! When the particle is inside the box, the potential energy of the system is constant.

3 Particle in a Box! We can define the potential energy to be infinitely large outside the box.! Because kinetic energy is always positive, to escape the box, the particle would need to have an infinite amount of energy.! The wave function for this problem can be expressed as: $ 2! x % " ( x) = Asin& # ' ( )! Applying the boundary conditions we get: # ( x) Asin n! " x $ = % L & ' (

4 Particle in a Box! The plots below show ψ versus x and ψ 2 versus x for n=1, 2, and 3.! Note that although ψ can be positive or negative, ψ 2 is always positive.! ψ 2 represents a probability density.

5 Particle in a Box! ψ 2 is zero at the boundaries, satisfying our boundary conditions.! In addition, ψ 2 is zero at other points. The number of zero points depends on the quantum number n.! Only certain wavelengths for particle are allowed.

6 Particle in a Box! The allowed wavelengths of the particle: λ=2l/n.! Because of this, the magnitude of the momentum of the particles is also restricted to specific values that can be found using the de Broglie wavelength: P = h/λ = h/(2l/n) = nh/(2l).! From this expression, we can find the allowed values of the energy (which is simply the kinetic energy of the particle since the potential energy is zero): E n = mv 2 /2 = p 2 /2m = (nh/2l) 2 /2m = = (h 2 /8mL 2 )n 2, where n = 1, 2, 3, P = 6.6! 10 "34 Js! n 2L! As we see from this expression, the energy of the particle is quantized.

7 ! The lowest allowed energy level corresponds to the ground state.! Other levels are called excited states. Particle in a Box 2! h " 2 # 1, 2, 3 2 $ En = n n = % 8mL &! n = 0 describes an empty box.! The particle can never be at rest if it is there at all.! The lowest energy is E 1.

8 Particle in a Box! Boundary conditions are applied to determine the allowed states of the system.! Boundary conditions describe interaction of the particle with its environment.! If this interaction restricts the particle to a finite region of space, it results in quantization of the energy of the system.! Because particles have wave-like characteristics, the allowed quantum states of a system are those in which the boundary conditions on the wave function representing the system are satisfied.

9 Summary of Ch. 28 Concepts Energy is quantized Absorption and emission of energy quanta is best thought of in terms of particles being absorbed or emitted. Collections of such particles, e.g. beams of light, electrons, etc. are described by wave functions that form interference patterns etc. These interference patterns describe the probability density for finding a particle in a given location. We can thus use what we learned about waves and boundary conditions in order to calculate the wave function and thus determine these probability densities.

10 End of Chapter 28

Atomic Physics! Quantum mechanics eventually resolved how the subatomic world works.! However, on its way to resolution, some amusing ideas were pondered, and ruled out.

11 Atomic Physics! Quantum mechanics eventually resolved how the subatomic world works.! However, on its way to resolution, some amusing ideas were pondered, and ruled out.! One of the theories ruled out by experiment was the Plum Pudding Model by J.J. Thomson.! In this model the atom was thought to be a large volume of positive charge with smaller electrons embedded throughout.! Almost like a watermelon with seeds.

12 ! A few alpha particles were deflected from their original paths (some even reversed direction). Atomic Physics! To test this model, Ernest Rutherford shot a beam of positively charged particles (alphas) against a thin metal foil.! Most of the alpha particles passed directly through the foil.

13 He then predicted that the electrons would orbit the nucleus like planets orbit the sun. Atomic Physics This thin foil experiment led Rutherford to believe that positive charge is concentrated in the center of the atom, which he called the nucleus. Centripetal acceleration should keep them from spiraling in (like the Moon). Thus, his model was named the Planetary Model.

14 ! It gives off light of a particular frequency.! As light is given, the electron will lose energy and its radius should decrease.! The electron should eventually spiral into the nucleus. Atomic Physics! But there were a few problems with the Planetary model of the atom. Such as, what happens when a charged particle (like the electron) is accelerated?

15 Emission Spectra Finally, the key to understanding atoms was to look at the light that was emitted from them. When a low-pressure gas is subjected to an electric discharge, it will emit light characteristic of the gas. When the emitted light is analyzed with a spectrometer, we observe a series of discrete lines.

16 Emission Spectra! This is known as emission spectra.! Each line has a different wavelength (color).! The elemental composition of the gas will tell what the resulting color lines will be.! Note that in general elements with a higher atomic number will have more lines.

17 Emission Spectra! The easiest gas to analyze is hydrogen gas.! Four prominent visible lines were observed, as well as several ultraviolet lines.! In 1885, Johann Balmer, found a simple functional form to describe all of the observed wavelengths:! where RH is known as the Rydberg constant RH = x10 7 m -1.! n = 3, 4, 5,...

18 Emission Spectra! Every value of n led to a different line in the spectrum.! For example, n = 3 led to a λ 3 = 656nm and n = 4 led to a λ 4 = 486nm.! The series of lines described by this equation is known as the Balmer Series.! Note how the spacing between the lines gets closer and closer the smaller the wavelength gets.

19 Absorption Spectra In addition to emission spectra (lines emitted from a gas), there is also absorption spectra (lines absorbed by a gas). An element can also absorb light at specific wavelengths. An absorption spectrum can be obtained by passing a continuous radiation spectrum through a cloud of gas. The elements in gas will absorb certain wavelengths.

20 Absorption Spectra! The absorption spectrum consists of a series of dark lines superimposed on an otherwise continuous spectrum.! The dark lines of the absorption spectrum coincide with the bright lines of the emission spectrum.! This is how the element of Helium was discovered.

21 Hydrogen Atom! In 1913 (2 years after the Rutherford experiment), Neils Bohr explained atomic spectra by utilizing Rutherford s Planetary model and quantization. In Bohr s theory for the hydrogen atom, the electron moves in circular orbit around the proton. The Coulomb force provides the centripetal acceleration for continued motion.

22 Hydrogen Atom! Only certain electron orbits are stable.! In these orbits the atom does not emit energy in the form of electromagnetic radiation.! Radiation is only emitted by the atom when the electron jumps between stable orbits.

23 Hydrogen Atom! The electron will move from a more energetic initial state to less energetic final state.! The frequency of the photon emitted in the jump is related to the change in the atom s energy:! If the electron is not jumping between allowed orbitals, then the energy of the atom remains constant.

24 Hydrogen Atom! Bohr then turned to conservation of energy of the atom in order to determine the allowed electron orbitals.! The total energy of the atom will be: =! 1 2 k e e 2 r! But the electron is undergoing centripetal acceleration (Newton s second law):

25 How do we quantize r?! Recall from classical mechanics that there was this variable known as angular momentum, L.! Angular momentum, L, was defined as: L = I ω! where I was rotational inertia and ω was angular velocity.! For an electron orbiting a nucleus we have that:! Giving us:

26 Hydrogen Atom! Bohr postulated that the electron s orbital angular momentum must be quantized as well:! where ħ is defined to be h/2π.! This gives us a velocity of:! Substituting into the last equation from two slides before:

27 Hydrogen Atom! Solving for the radii of Bohr s orbits gives us:! The integer values of n = 1, 2, 3, give you the quantized Bohr orbits.! Electrons can only exist in certain allowed orbits determined by the integer n.! When n = 1, the orbit has the smallest radius, called the Bohr radius, ao.! ao = nm

28 Hydrogen Atom: Bohr s Theory We know that the radii of the Bohr orbits in a hydrogen tom are quantized: We also know that when n = 1, the radius of that orbit s called the Bohr radius (ao = nm). So, in general we have: r n = n 2! 2 m e k e e 2 rn = n 2 ao The total energy of the atom can be expressed as : E = " 1 # tot % 2 k e $ e 2 r & ( ' (assuming the nucleus is at rest)

29 Hydrogen Atom E = " 1 # tot % 2 k e $ e 2 & n 2 ( = a o ' "13.6 ev n 2! This is the energy of any quantum state (orbit). Please note the negative sign in the equation.! When n = 1, the total energy is 13.6eV.! This is the lowest energy state and it is called the ground state.! The ionization energy is the energy needed to completely remove the electron from the atom.! The ionization energy for hydrogen is 13.6eV.

30 ! So, a general expression for the radius of any orbit in a hydrogen atom is: rn = n 2 ao! The energy of any orbit is: Hydrogen Atom! If you would like to completely remove the electron from the atom it requires 13.6eV of energy.

31 Hydrogen Atom! What are the first four energy levels for the hydrogen atom?! When n = 1 => E1 = 13.6eV.! When n = 2 => E2 = 13.6eV/2 2 = 3.40eV.! When n = 3 => E3 = 13.6eV/3 2 = 1.51eV.! When n = 4 => E4 = 13.6eV/4 2 = 0.850eV.! Note that the energy levels get closer together as n increases (similar to how the wavelengths got closer in atomic spectra).! When the atom releases a photon it will experience a transition from an initial higher energy level (ni) to a final lower energy level (nf).

32 ! The energies can be compiled in an energy level diagram.! As the atom is in a higher energy state and moves to a lower energy state it will release energy (in the form of a photon).! The wavelength of this photon will be determined by the starting and ending energy levels. Hydrogen Atom

33 Hydrogen Atom! The photon will have a wavelength λ and a frequency f: f = E i " E f h! To find the wavelengths for an arbitrary transition from one orbit with nf to another orbit with ni, we can generalize Rydberg s formula: 1 ( = R H & 1 $ % n 2 f ' 1 n 2 i #! "

34 Hydrogen Atom! The wavelength will be represented by a different series depending on your final energy level (nf).! For nf = 1 it is called the Lyman series (ni = 2,3,4,...).! For nf = 2 it is called the Balmer series (ni = 3,4,5...).! For nf = 3 it is called the Paschen series (ni = 4,5,...).

35 Concept Question! When a cool gas is placed between a glowing wire filament source and a diffraction grating, the resultant spectrum from the grating is which one of the following?! A) line emission.! B) line absorption.! C) continuous.! D) monochromatic.! E) de Broglie.

36 Atomic Spectra! Example! What are the first four wavelengths for the Lyman, Balmer, and Paschen series for the hydrogen atom?! Answer! The final energy level for either series will be nf = 1 (Lyman), nf = 2 (Balmer), and nf = 3 (Paschen).

37 ! Answer Atomic Spectra! Turn to the generalized Rydberg equation: 1 (! For the Lyman series we have: = R H & 1 $ % n 2 f ' 1 n 1 " = R $ 1 H 1 # 1 ' $ 2 n & 2 ) = R i H & 2 % n i ( n # 1 2 % i n i 2 i #! " ' ) ( 1 " = R H $ n 2 #1' & i n 2 % i ( ) " n = 1 R H $ & % 2 n i ' ) = n 2 i #1( 1 $ 2 n i ' & ) *10 7 m -1 % n 2 i #1(

38 ! Answer Atomic Spectra! Finally for the Lyman series: 1 % 2 2 ( " 1 = ' * =121 nm #10 7 m -1 & 2 2 $1) " 2 = " 3 = " 4 = 1 % 3 2 ( ' * =103 nm #10 7 m -1 & 3 2 $1) 1 % 4 2 ( ' * = 97.2 nm #10 7 m -1 & 4 2 $1) 1 % 5 2 ( ' * = 95.0 nm #10 7 m -1 & 5 2 $1)! For the Balmer series we have from before:! λ1 = 656nm, λ2 = 486nm, λ3 = 434nm, λ4 = 410nm.

39 Atomic Spectra! Answer! For the Paschen series we have: 1 " = R H $ & % 1 9 # 1 n i 2 ' $ 2 n ) = R i H & 2 ( 9n # 9 % i 9n i 2 ' ) ( " n = 1 % 2 9n i ( ' * #10 7 m -1 & n 2 i $ 9) " 1 = 1 % 9( 2 4 ) ( #10 7 m -1 ' & 4 2 $ 9* =1880 nm " = 1 % 9( 2 5 ) ( 2 ) #10 7 m -1 ' & 5 2 $ 9* ) =1280 nm " 3 = 1 % 9( 2 6 ) ( #10 7 m -1 ' & 6 2 $ 9* =1090 nm " = 1 % 9( 2 7 ) ( 4 ) #10 7 m -1 ' & 7 2 $ 9* =1010 nm )

40 Atomic Spectra! The only series that lies in the visible range ( nm) is the Balmer series.! The Lyman series lies in the ultraviolet range and the Paschen series lies in the infrared range.! We can extend the Bohr hydrogen atom to fully describe atoms that are close to hydrogen.! These hydrogen-like atoms are those that only contain one electron. Examples: He+, Li++, Be+++! In those cases, when you have Z as the atomic number of the element (Z is the number of protons in the atom), you replace e 2 with Ze 2 in the hydrogen equations.

41 Concept Question! Consider a hydrogen atom, a singly-ionized helium atom, a doubly-ionized lithium atom, and a triplyionized beryllium atom. Which atom has the lowest ionization energy?! A) hydrogen! B) helium E = " 1 # tot % 2 k e $ e 2 r & ( r = n 2! 2 n ' m k e 2 e e! C) lithium! D) beryllium! E) the ionization energy is the same for all four

42 For Next Time (FNT)! Finish reading Chapter 29

Physics 1C. Lecture 28D

Physics 1C. Lecture 28D Physics 1C Lecture 28D "I ask you to look both ways. For the road to a knowledge of the stars leads through the atom; and important knowledge of the atom has been reached through the stars." --Sir Arthur