Physics 102: Lecture 26. X-rays. Make sure your grade book entries are correct. Physics 102: Lecture 26, Slide 1

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Transcription:

Physics 102: Lecture 26 X-rays Make sure your grade book entries are correct. Physics 102: Lecture 26, Slide 1

But first a quick review of the periodic table http://www.youtube.com/watch?v=smwlzwgmmwc http://www.youtube.com/watch?v=gfivxvmbi I0&feature=related Physics 102: Lecture 26, Slide 2

X-Rays Photons with energy in approx range 100eV to 100,000eV. This large energy means they go right through you (except for your bones). What are the wavelengths?.01 nm to 10 nm λ= hc 1240 ev nm = E E 1240 ev nm 100000 ev.01nm 1240 100 Physics 102: Lecture 26, Slide 3 10 nm

Physics 102: Lecture 26, Slide 4

X-Ray Production How do you produce 100 ev photons? Black Body Radiation Would require temperature over 10 times hotter than surface of sun Excitation of outer electrons Typically have energy around 10 ev Radioactive Decays Hard to turn on/off Physics 102: Lecture 26, Slide 5

Electron Tubes Accelerate an electron through a voltage difference to give it some energy... An electron is accelerated through a potential difference of 70,000 V. How much energy does it emerge with? Recall from Lecture 3: EPE = V q KE = EPE = (70,000 V) (1 e - ) = 1.6 x 10-19 C = 70,000 ev = 11.2 x 10-14 J EPE of voltage gap becomes K.E. for electron. Physics 102: Lecture 26, Slide 6

From Electrons to X-Rays Now take these high energy electrons (up to 100,000 ev) and slam them into heavy atoms - any element. 2 kinds of X-Rays are produced: VIS Bremsstrahlung Characteristic Physics 102: Lecture 26, Slide 7

Physics 102: Lecture 26, Slide 8

Bremsstrahlung X-Rays Electron hits atom and slows down, losing kinetic energy. Energy emitted as photon Electron hitting atom makes many photons (X-Rays), all with different energy. Many different wavelengths. intensity If all of electron s energy is lost to a single photon, photon has maximum energy (minimum wavelength). Minimum X-Ray wavelength = λ o. Physics 102: Lecture 26, Slide 9 λ 0 λ

Bremsstrahlung Practice An electron is accelerated through 50,000 volts What is the minimum wavelength photon it can produce when striking a target? Minimum wavelength Maximum energy Electron loses ALL of its energy in one collision and emits one photon. intensity λ 0 = hc E = 1240 50,000 Physics 102: Lecture 26, Slide 10 =.0248 nm λ 0 λ

Characteristic X-Rays Electron knocks one of the two K shell (ground state) electrons out of an atom. L (n=2) or higher shell electron falls down to K shell (ground state) and x-ray photon is emitted e - e - e - L shell (n=2) Characteristic x-ray nomenclature e - Physics 102: Lecture 26, Slide 11 e - e - (high energy electron) K shell (n=1) n=1 K shell n=2 L shell n=3 M shell

Physics 102: Lecture 26, Slide 12

Characteristic X-Rays Electron knocks one of the two K shell (ground state) electrons out of an atom. L (n=2) or higher shell electron falls down to K shell (ground state) and x-ray photon is emitted ejected electron e - e - e - e - e - e - e - L shell (n=2) K shell (n=1) Characteristic x-ray nomenclature n=1 K shell n=2 L shell n=3 M shell Physics 102: Lecture 26, Slide 13

Characteristic X-Rays Electron knocks one of the two K shell (ground state) electrons out of an atom. L (n=2) or higher shell electron falls down to K shell (ground state) and x-ray photon is emitted e - e - e - e - L shell (n=2) L shell electron falls down e - K shell (n=1) e - Characteristic x-ray nomenclature n=1 K shell n=2 L shell n=3 M shell X-Ray photon emitted Physics 102: Lecture 26, Slide 14 K α X-ray (n=2 n=1 transition)

photon K α X-Rays Estimate the energy of K α X-rays off of a silver (Ag) target (Z=47). Better formula for Careful! the formula n = ( 13.6)Z2 2 ( 13.6)( Z 1) multi-electron atoms En = 2 n 2 n=2 L assumed a single electron bound to just a positive nucleus. E n=1 L = 13.6eV(47 1) 2 1 K 2 = 7.2keV 2 E K = 13.6eV(47 1) 2 1 1 = 28.8keV 2 E(K α ) = E L E K = 21.6 kev (vs. 21.7 kev Expt) Not bad! Physics 102: Lecture 26, Slide 15 intensity K α λ

Physics 102: Lecture 26, Slide 16

K β X-Rays K α X-rays come from n=2 n=1 transition. What about n=3 n=1 transition? Not as likely, but possible. Produces K β X-Rays! K β X-Rays are higher energy (lower λ) than K α. (and lower intensity) intensity K β K α λ Different elements have different Characteristic X-Rays Physics 102: Lecture 26, Slide 17

All Together Now... Brehmsstrahlung X-Rays and Characteristic X-Rays both occur at the same time. intensity λ 0 λ intensity K β K α λ intensity K β K α λ λ 0 Physics 102: Lecture 26, Slide 18

Preflight 26.1 K α intensity intensity K β K β K α λ λ These two plots correspond to X-Ray tubes that: (1) Are operating at different voltages (2) Contain different elements (3) Both (4) Neither Physics 102: Lecture 26, Slide 19

Physics 102: Lecture 26, Slide 20

ACT: X-Rays I intensity intensity λ Which graph corresponds to the tube being operated at the higher voltage? 1) Top 2) Bottom Physics 102: Lecture 26, Slide 21 λ

ACT: X-Rays II intensity intensity λ The top spectrum comes from a tube with a silver target (Ag, Z=47). What is the bottom target? 1) Pd, Z=46 2) Ag, Z=47 3) Cd, Z=48 Physics 102: Lecture 26, Slide 22 λ

From atoms to nuclei to nucleons to quarks: The heirarchy of sizes Physics 102: Lecture 26, Slide 23

Physics 102: Lecture 26, Slide 24

Nuclear Physics A Z 6 3 Li Nucleus = Protons+ Neutrons nucleons Z = proton number (atomic number) Gives chemical properties (and name) N = neutron number A = nucleon number (atomic mass number) Gives you mass density of element Physics 102: Lecture 26, Slide 25 A=N+Z Periodic_Table

Preflight 27.1 A material is known to be an isotope of lead Based on this information which of the following can you specify? 1) The atomic mass number 2) The neutron number 3) The number of protons Physics 102: Lecture 26, Slide 26

Strong Nuclear Force Rutherford experiment shows that all the positive charge is contained in a small nucleus Size ~ few x 10-15 m (few fm) Let s estimate EPE of two protons separated by 1 fm EPE = kq 2 /r = (9 x 10 9 )(1.6 x 10-19 ) 2 /10-15 = 2.3 x 10-13 J = 1.44 x 10 6 ev = 1.44 MeV Therefore, the force that binds protons and neutrons together to form a nucleus must be very strong in order to overcome Coulomb repulsion But the force acts over very short distances of order few fm Two atoms don t feel force Physics 102: Lecture 26, Slide 27

Physics 102: Lecture 26, Slide 28

Strong Nuclear Force Hydrogen atom: Binding energy =13.6eV (of electron to nucleus) Coulomb force proton electron neutron Simplest Nucleus: Deuteron=neutron+proton proton Very strong force Binding energy of deuteron = 2.2 10 6 ev or 2.2Mev! That s around 200,000 times bigger! Physics 102: Lecture 26, Slide 29

# protons = # neutrons Pauli Principle - neutrons and protons have spin like electron, and thus m s = ±1/2. n n p p n n p p Can get 4 nucleons into n=1 state. Energy will favor N=Z But protons repel one another (Coulomb Force) and when Z is large it becomes harder to put more protons into a nucleus without adding even more neutrons to provide more of the Strong Force. For this reason, in heavier nuclei N>Z. Physics 102: Lecture 26, Slide 30 7

Nuclei have energy level just like atoms 60 Ni energy levels 12 C energy levels Note the energy scale is MeV rather than ev Physics 102: Lecture 26, Slide 31

Physics 102: Lecture 26, Slide 32