Bohr model and Franck-Hertz experiment

Bohr model and Franck-Hertz experiment Announcements: Will finish up material in Chapter 5. There will be no class on Friday, Oct. 18. Will announce again! Still have a few midterms see me if you haven t picked it up Today will be analyzing the Bohr model with clicker questions. Gustav Hertz: 1887-1985 James Franck: 1882-1964 http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 1

Clicker question 1 What atomic energy levels for electrons are consistent with this spectrum? Electron Energy levels: 0 ev 0 ev 5ev 3ev 2ev Photon energy 0 100 200 300 400 500 600 700 800 nm A B C D 0 ev 10 ev E 2 ev 3 ev 7 ev 8 ev 10 ev 7 ev 10 ev 7 ev 5 ev 0 ev 5 ev 3 ev 2 ev 0 ev http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 2

Why are only certain energy levels allowed? Bohr postulated that angular momentum was quantized so: Quantizing angular momentum leads to a quantization of radius: where where Quantizing radius leads to a quantization of energy: is the Bohr radius where E R is the Rydberg energy: http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 4

Last lecture we found - + v r If the electron orbits the proton at a constant speed, the magnitude of the net force on the electron is mv 2 /r Setting the net force (from Coulomb) equal to the mass times acceleration (mv 2 /r) for circular motion gives us: which we can also write as: http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 5

What does this say about total energy? - r F + v We now put together three pieces: 1. mv 2 = ke 2 /r (just derived) 2. electrostatic potential energy is U = -ke 2 /r 3. nonrelativistic kinetic energy is K = ½mv 2 This means K = -½U and the total energy is 0 potential energy distance from proton Note E = -K and 2K = -U The total energy, radius, and velocity are all related. Knowing just one of the three determines the other two! http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 6

A little algebra mvr =nħ = nh/2π, n = 1, 2, 3,.. So r = n 2 ħ 2 /m ke 2, n = 1, 2, 3,. r =.053 nm for n =1 And v = ke 2 /nħ, v = 2.2 10 6 m/s n = 1, 2, 3,. Justifies using classical mechanics, but for large Z need to be careful E = -K = mv 2 /2 = -mk 2 e 4 /2n 2 ħ 2, n = 1, 2, 3,. http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 7

Clicker question 2 Which of the following principles of classical physics is violated in deriving the Bohr model of the atom? A. Opposite charges attract with a force inversely proportional to the square of the distance between them. B. The force on an object is equal to its mass times its acceleration. C. Accelerating charges radiate energy. D. Particles have a well-defined position and momentum. E. All of the above. http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 8

Clicker question 2 Which of the following principles of classical physics is violated in deriving the Bohr model of the atom? A. Opposite charges attract with a force inversely proportional to the square of the distance between them. B. The force on an object is equal to its mass times its acceleration. C. Accelerating charges radiate energy. D. Particles have a well-defined position and momentum. E. All of the above. Note that A & B were used in the derivation of the Bohr model. http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 9

Bohr s calculation of hydrogen energy levels Using the formula for energy Hydrogen energy levels Bohr could calculate the various transitions and they agreed with the generalized Balmer formula. Photon energy is given by: E γ = E n E n' = E R 1 n' 2 1 n 2 = 13.6 ev 1 n' 2 1 n 2 http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 10

Clicker question 3 An atom with the energy levels shown is initially in the ground state. A free electron with an energy of 16.0 ev hits the atom. What possible states could the atom be in after the interaction? A. n=1 B. n=1, n=2, or n=3 C. n=3 D. n=2 or n=3 E. Any of the states n=5 n=4 n=3 n=2 n=1 1 ev 2 ev 10 ev 20 ev http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 11

Clicker question 3 An atom with the energy levels shown is initially in the ground state. A free electron with an energy of 16.0 ev hits the atom. What possible states could the atom be in after the interaction? A. n=1 B. n=1, n=2, or n=3 C. n=3 D. n=2 or n=3 E. Any of the states n=5 n=4 n=3 n=2 n=1 If the atom goes to n=1, 2, or 3, the free electron will lose 0 ev, 10 ev, or 15 ev of kinetic energy. 1 ev 2 ev 10 ev 20 ev Free electron hits atom e e e Less KE http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 12

Clicker question 4 An atom with the energy levels shown is initially in the ground state. A photon with an energy of 16.0 ev hits the atom. What possible states could the atom be in after the interaction? A. n=1 B. n=1, n=2, or n=3 C. n=3 D. n=2 or n=3 E. Any of the states n=5 n=4 n=3 n=2 n=1 1 ev 2 ev 10 ev 20 ev For the free electron, whatever energy is absorbed by the atom is deducted from the free electron s kinetic energy. For photons, the photon is absorbed and so it must transfer all of its energy to the atom. Otherwise energy would not be conserved. Therefore, atoms can only absorb photons with an energy that will exactly move the atom to another energy level. http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 14

Clicker question 5 An atom with the energy levels shown is initially in the ground state. Which of the following is the most complete list of photon energies that can be absorbed by the atom? A. 1 ev B. 1 ev, 2 ev, 5 ev, 10 ev C. 10 ev D. 10 ev, 15 ev, 18 ev E. 10 ev, 15 ev, 18 ev, 25 ev n=5 n=4 n=3 n=2 n=1 1 ev 2 ev 10 ev 20 ev Photons with energy of 10 ev, 15 ev, 18 ev, 19 ev will cause the electron to jump to the n=2, n=3, n=4, n=5 energy level. Any photon with an energy 20 ev will ionize the atom. The electron will escape. Since a free electron can have any energy, any photon with energy 20 ev can be absorbed by the atom. 20 ev will go into ejecting the electron and the rest will go into the free electron s kinetic energy. Similar to the photoelectric effect. http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 16

Hydrogen like ions Atoms which have only one electron can be analyzed much like the hydrogen atom. An atom with atomic number Z with Z-1 electrons removed is a hydrogen like ion The (Coulomb) force on the electron is The increase in the force results in tighter orbits and a deeper potential well, reducing the energy (more negative). http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 17

Clicker question 6 A single electron is in the n=2 energy level around a helium nucleus (He + ). What is the minimum energy photon that can remove this electron? A. 3.4 ev B. 13.6 ev C. 27.2 ev D. 54.4 ev E. None of the above http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 18

Clicker question 6 electron? A. 3.4 ev B. 13.6 ev C. 27.2 ev D. 54.4 ev E. None of the above A single electron is in the n=2 energy level around a helium nucleus (He + ). What is the minimum energy photon that can remove this The atomic number of helium is Z=2 and the electron is in the n=2 energy level. So the energy of the state is The value E n is also referred to as the binding energy since it is a measure of how bound the electron is. It takes that amount of energy to free the electron (break its bond to the proton). http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 19

Reduced Mass Have shown that energy for one electron atoms levels scale as E = -K = mv 2 /2 = -mk 2 Z 2 e 4 /2n 2 ħ 2, n = 1, 2, 3,. But we have one more correction it was found that the energy levels in Hydrogen were all lower by 1 part out of 1800. This was accounted for by the reduced mass, because we had assumed the nucleus was infinitely heavy L = m e M N /(m e +M N ) x vr = nħ Reduced Mass http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 20

E = 1 2 m e v 2 E = 1 2 m e r 2 Reduced Mass + V(r) + V(r) E = 1 2 m er 2 1 + 1 2 M r 2 2 + V(r 2 r 1 ) Plugging back into energy equation We find: E = 1 2 m e M 2 (m e + M) 2 r 2 21 { + 1 2 2 M m e (m e + M) 2 r 2 21 r 21 = r 2 r 1 m e r 1 + M r 2 = 0 r 1 = r 2 = M m e + M m e m e + M r 21 r 21 + V(r 2 r 1 ) E = 1 2 m e M r (m e + M) 2 21 + V(r 2 r 1 ) http://www.colorado.edu/physics/phys2170/ Physics 2170 Fall 2013 21