There are two ways to numerically solve this. Both pay attention to the fact that kev is not a standard metric units. 2 45!"# 3.00!10!!/! 1 !

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Angelina Kennedy
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1 Physics 49 Midterm Exam Solutions Oct th 0 ) An electron of kinetic energy 45keV moves in a circular orbit perpendicular to a magnetic field of 0.35T. Find the radius of the orbit in meters (5 points) "# There are two ways to numerically solve this. Both pay attention to the fact that kev is not a standard metric units. 45"# / " 5"# " ".76x0 C/kg 0.35.x0 In this case I canceled out the kev units leaving only metric units. " " " " " " " 45"#.76x0 C/kg 0.35.x0 In this case I use the definition of electron volts. ev is the energy of an electron accelerated though a potential difference of V. The units are electron-charge*volts. By dividing by the electron charge you convert to Volts, which is a standard metric unit. ) Derive the units of the fine structure constant. (5 points) ℏ There are various easy ways to do this. ke is the top portion of the Electric force equation for a point charge, F ke/r (covered in Rutherford scattering). ke Fr [(force*distance)*distance] [Energy*distance] hc: energy*distance. Therefore alpha is dimensionless. A second method The potential of an electric field (not potential energy) is: [""#] "#$%&'( ["#$% "#$%&'(] hc: energy*distance

2 Therefore alpha is dimensionless. (this methods puts the numerator and denominator in the same energy units) The fact that alpha is dimensionless is often cited as a reason why it a truly fundamental constant. 3) Demonstrate that the wave function of the lowest energy level of the particle in a box obeys the uncertainty relation by calculating Δ and Δp using expectation values. For the potential use: V(x) 0, 0 < x < L V(x) infinity x < 0, x > L (30 points) The wave function must satisfy the Schrodinger equation inside the box and be zero at x0 and xl The general formula for the wave function is: "# from x0 to xl The normalization can be found from normalizing the n ground state wave function. (No point taken off is the student starts from this wave function.) The lowest energy level should result in a minimum Δ and Δp consistent with the uncertainty principle. "# The uncertainty principle is defined by multiplying the standard deviation of x and p. (chapter 5.5) (The definition of the standard deviation from the mean was defined in the lecture on expectation values): Note: 3 of the 4 integrals we to perform to evaluate these expectation values have trivial solutions. I will evaluate of 4 since the easy solution to the 3 rd integral is not so obvious. +,, " "# "# " "# " using 6 " 4 8 "# 4 "#

3 6 4 3 The wave function is symmetric and centered at L/ so: 4, ℏ "# ℏ "#, " ℏ "# " " using ℏ ℏ Note you also get from first energy level. Since energy is a constant proportional to and " ". The momentum has equal probability to be positive or negative so on average is zero: 0 ΔΔ ℏ ℏ ℏ 0.568ℏ > ℏ

4 Which obeys the uncertainty relationship and matches the expectation that the lowest energy level will have the smallest uncertainty since the uncertainty in momentum is the smallest. 4) Explain why the infinite potential well wave function satisfied the uncertainty principle, but with slightly larger uncertainty in x and p, while the lowest energy state of the harmonic oscillator satisfies the uncertainty principle with the minimum possible uncertainty in x and p. (5 points) The lowest energy state of the harmonic oscillator has a Gaussian probability distribution in space, which is the same probability distribution for the shape of wave packet. For the wave packet (or quantum harmonic oscillator) when you calculate the distributions of momentums you also find a Gaussian distribution. As explained in the text book the combination of Gaussian distributions in both space and momentum for the wave packet results in the minimum possible uncertainty. Any other probability distribution in space will have a larger uncertainty. The first cos function solution is similar to a Gaussian in that it is peaked in the center of the potential and falls off to either side to but not exactly the same resulting in a slightly larger uncertainty. Another argument that can be made is that any system that highly localizes x or p, in this case x is highly constrained to be inside the box, tends to make the other value, p or x respectively, have higher uncertainty driving the uncertainty over the minimum value. 5) For an electron in a finite potential well with L0.00nm and V0MeV how many bound state solutions are there. The conditions for the quantized energy solutions are: tan and cot (5 points) recast this as tan ħ and cot ħ together tan and cot will have zeros at including n0 where the st tangent solution starts. and any tan or cot line starting there will cross the parabolic curve if the maximum of the parabolic curve is larger. the maximum solution is at

5 ħ 4 " h 0.5"# 000" 0"# 40"#$% bound states exist.

Summary of Last Time Barrier Potential/Tunneling Case I: E<V 0 Describes alpha-decay (Details are in the lecture note; go over it yourself!!) Case II:

Quantum Mechanics and Atomic Physics Lecture 8: Scattering & Operators and Expectation Values http://www.physics.rutgers.edu/ugrad/361 Prof. Sean Oh Summary of Last Time Barrier Potential/Tunneling Case