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Midterm Exam Solutions

Midterm Exam Solutions. Problem 1: Of a certain kind of seed, 25% normally germinates. To test a new germination stimulant , 100 of these seeds are planted and treated with the stimulant. If 32 of them germinate , can you conclude (at the 5% level of significance) that the stimulant helps ?

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Midterm Exam Solutions

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  1. Midterm Exam Solutions • Problem 1: Of a certain kind of seed, 25% normally germinates. To test a new germination stimulant, 100 of these seeds are planted and treated with the stimulant. If 32 of them germinate, can you conclude (at the 5% level of significance) that the stimulant helps? • This was a homework problem. The method that you should use is the Binomial Distribution or the Gaussian Approximation to it. • = 0.024 • = 0.069 • The Poisson Distribution is when n is very large and p is very small. In this case the Binomial is indistinguishable from the Poisson.

  2. Midterm Exam Solutions • Multiple Choice Problem 1: • Below is a container and the numbers correspond to the surface numbers. Which below describes the interior of the surface? • a. -1 4 -5 2 3 • b. 1 -4 5 -2 -3 • c. 1 -4 5 -2 : -3 • d. 1 -4 5 (-2 : -3) • e. 1 -4 -5 -2 : -3

  3. Midterm Exam Solutions • Multiple Choice Problem 2: • If you have a 3 MeV alpha particle and a 5 MeV alpha particle how many alpha particles do you have? • a. 3 • b. 2 • c. 5 • d. 8 • e. There are no such thing as alpha particles

  4. Midterm Exam Solutions • Multiple Choice Problem 3: • Use step-by-step propagation to determine the final value and the uncertainty in the following (20 ± 1)/[ (5.0 ± 0.1)- (3.0 ± 0.1)] • a. 10 ± 1 • b. 10 ± 0.7 • c. 10.0 ± 1.0 • d. 10.0 ± 0.1 • e. 10.0 ± 0.6

  5. Midterm Exam Solutions • Multiple Choice Problem 3: • 0.866 = 1

  6. Midterm Exam Solutions • Multiple Choice Problem 3: • Denominator = • Quotient of numerator and denominator • = fractional uncertainty = 0.086 • Total uncertainty is 0.86 = 1

  7. Midterm Exam Solutions • Multiple Choice Problem 4: • Which of the following statements are false? • a. Protons and electrons are charged and gamma rays and neutrons are uncharged • b. Protons, electrons, and gamma rays are charged and neutrons are uncharged • c. Neutrons interact with nuclei while protons, electrons and gamma rays interact with electrons • d. Stopping power is a measure of energy loss or a particle per unit time • e. Gamma rays nearly always emitted after a radioactive decay process • f. Unstable isotopes can decay purely through gamma-ray emission • g. It is physically not possible for heavy charged particles to create Bremsstrahlung radiation • h. Heavy charged particles are defined as alpha particles and heavier • i. Alpha particles travel in a straight line through matter nearly all the time • j. Gamma rays travel in a straight line through matter nearly all the time

  8. Midterm Exam Solutions • Multiple Choice Problem 5: • Where is the fire extinguisher? • a. In Dr. Chvala’s office • b. In the back of the lab • c. In the hallway by the entrance • d. There is no fire extinguisher • e. In the safe

  9. Lab This Week • Will be testing SCAs and MCAs • It is important that each student is familiar with how each of these work • This lab will require the use of • Oscilloscope • Tail pulse generator • Amplifier • SCA • MCA • Maestro software on the PC in the lab

  10. Connector types

  11. Connector types

  12. Tail Pulse Generator • Generate a pulse with a controlled rise and fall time • Can control features like • Frequency • Amplitude • Shape • Rise time • Fall time • Pulse polarity

  13. Amplifier • Used to amplify and shape a pulse • Coarse gain • Fine gain • Shaping time • Polarity • Output

  14. Oscilloscope

  15. SCA • Normal • Differential SCA • INT • Integral SCA • Window • Differential mode of the SCA

  16. MCA • Two inputs • Two gates • The gates are used to control when the MCA will accept pulses

  17. Purpose of the Lab • Will be investigating how an SCA works • Will be investigating how an MCA responds to various pulses • Will test the MCA dead time

  18. NE 401 Spring 2013 Grade Distribution

  19. NE 401 Spring 2013 Grade Distribution

  20. Semiconductors I NE 401, Spring 2013

  21. Scintillation detectors • Inefficient process of light creation • ~100 eV or more required per carrier • Statistical variation of small number of charge carriers leads to an intrinsic poor energy resolution • So what can we do to increase the energy resolution?

  22. Semiconductor Detectors • Density ~103 times greater than a gas • High intrinsic efficiency • Suited for high energy electron and gamma-ray detection • Fast timing • Limited size • High susceptibility to radiation damage

  23. Schrodinger’s Wave Equation • Time dependent • Time independent • The wave function is a probability density function so we require that

  24. Particle in a Box

  25. Particle in a Box • In regions I and III the wave function must be zero • In region two we have • The solution is • The boundary conditions are

  26. Particle in a Box • The condition is true only when • Further, the probability must be one where it exists, so • Therefore, our solution is

  27. Particle in a Box • Remember that and • This means that • This means that the particle energies are quantized. • Particles in this box can only have a fixed energy

  28. The One Electron Atom • Schrodinger's equation is • The potential is • An atom exists in 3D space and it is a good idea to take advantage of spherical symmetry by transforming to spherical-polar coordinates

  29. The One Electron Atom • Schrodinger's equation now becomes • Solve through separation of variables • Going through the solution you will find that

  30. The One Electron Atom • The solution provides a wave function dependent on three quantum numbers • n = Principle Quantum Number ( n ≥ 1 ) • l = Orbital angular momentum quantum number ( 0 ≤ l ≤ n-1 ) • m = Magnetic quantum number ( -l ≤ m ≤ l )

  31. The One Electron Atom

  32. Pauli Exclusion Principle • Electrons in a given system (atom, molecule, etc.) cannot occupy the same quantum state • This means the same set of quantum numbers • This means the same energy

  33. Pauli Exclusion Principle

  34. Pauli Exclusion Principle • What does this principle mean when we are in a solid?

  35. Pauli Exclusion Principle • The electron energy, or allowed energy states, will create a “band” of allowable energies at the stable equilibrium interatomic distance • Very small energy difference • Quasi-continuous energy distribution

  36. Pauli Exclusion Principle • When we have multiple occupied principle quantum energy states several bands are made • These bands represent allowed and forbidden energy states • The structure of and filling of these bands with electrons is what arises to differences between solid with respect to their electronic properties

  37. Pauli Exclusion Principle • A simple example

  38. Electrical Conduction in Solids • Complex example • Silicon with its four valence electrons • At T = 0o K • No thermal energy • At T > 0o K • Thermal energy

  39. Electrical Conduction in Solids • Depending on the band gap energy between energy states thermally excited electrons may cross the “forbidden” region into a higher energy band through breaking a covalent bond • Energy levels are so close that electrical conduction is “typically” possible while that electron stays in this excited state • Requires an applied electric field (voltage) and is called a drift current.

  40. Electrical Conduction in Solids Think of the left as physical space and the right as “energy space.” This means that the electron is not necessarily raising in position, only in energy

  41. Concept of the Hole The positively charged vacant location where the electron resides can move. This indicates that we have two different types of charge carriers, electrons and holes.

  42. Electrical Conduction in Solids • From the figures it can be seen that conduction will increase as the number of allowed “free” states increases in a solid • A metal, or material with a high conductivity, will have a wide band that is filled half way or have overlapping bands (zero bandgap energy)

  43. Electrical Conduction in Solids • So the question is, what is the density of charge carriers in the solid? • To answer this question we move to a statistical analysis: • Fermi-Dirac Probability Function • Particles are treated as indistinguishable but only one particle is allowed per energy state (this is what is needed for this type of system)

  44. Electrical Conduction in Solids • Fermi-Dirac Probability Function • EF is the Fermi energy

  45. Electrical Conduction in Solids • When E – EF >> kT this distribution becomes • Known as the Maxwell-Boltzmann approximation

  46. Thermal Excitation • As we can see as the temperature increases so does the probability of electron excitation

  47. Intrinsic Carrier Concentrations • Electron and hole concentrations

  48. Dopants in Semiconductors Extra electron has more weakly bound

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