Final Exam - Solutions PHYS/ECE Fall 2011

Transcription

1 Final Exam - Solutions PHYS/ECE 34 - Fall 211 Problem 1 Cosmic Rays The telescope array project in Millard County, UT can detect cosmic rays with energies up to E 1 2 ev. The cosmic rays are of unknown composition, but there is some evidence that they may be neutrons, which have a proper lifetime τ 885 s. If this is correct: (a) (5 points) What is the Lorentz factor, γ, for the high energy neutrons? E γmc 2 γ E m n c 2 (1) 1 2 ev ev (2) (3) So we see that these neutrons are extremely relativistic, and indeed we can approximate their velocity as v n c. (b) (5 points) The proper diameter of the earth is 12,56 km. What is the diameter of the earth along the direction of motion according to an observer moving with a neutron? In the neutron s reference frame, the earth is contracted along the direction of motion by the factor γ. Thus we have: L earth L earth γ 12, 56 km km.12 mm. (4) To the neutron, the earth is about a tenth of a millimeter in diameter! (c) (5 points) Given the limited time the neutrons exist before decaying into something else, what is the longest distance from the earth that they could have been produced? To an observer on earth, the lifetime of the neutrons is lengthened (dilated) by the Lorentz factor: τ γτ. In this time, the neutrons can travel a distance D n v n τ c γ τ, where we have used the approximation v n c, according to the answer in part (a) above. (Note: you can still get the right answer without this approximation, but it just takes much more computation.) Plugging in numbers, we get: D n (3 1 8 m/s)( )(885 s) m, (5) which corresponds to 3 1 light-years. Thus, the source of the high-energy neutrons cannot be (much) further away from the earth than this. For reference, the diameter of 1

2 2 the Milky Way galaxy is 1 5 light-years (source: Wikipedia), so the neutrons probably must come from the Milky Way or from a very nearby galaxy or cluster. This presents a problem because there are no known sources for such high-energy particles in the galaxy. Problem 2 Lorentz Transformation An electron initially at rest in the laboratory is accelerated by an electric potential difference of V 35 kv along the +y direction. A proton is moving in the laboratory at speed.5 c along the +x direction, making a right angle with direction +y. (a) (5 points) What is the 4-momentum of the accelerated electron in the laboratory frame? y e - p + Frame S x After acceleration, electrons move in the +y direction in the laboratory reference frame with kinetic energy, K e 35 kev. Thus, the total energy of the electrons in the laboratory frame is E e K e + m e c 2 ( ) kev 861 kev. In the lab frame, therefore, the electrons have 3-momentum (in energy units): p e c E 2 e m 2 ec 4 (861) 2 (511) 2 kev 693 kev, (6) directed along the +y direction. The 4-momentum can now be written down explicitly: E e /c 861 p x p y 693 kev/c. () p z (b) (5 points) Use a Lorentz transformation to find the 4-momentum of the accelerated electron from the proton s point of view. As we ve designated the laboratory as the S frame, the proton frame is then the S frame, which is obviously moving to the right with respect to the laboratory. Therefore, we just need to use the normal form of the Lorentz transformation matrix. The problem specifies that the S frame is moving with β v/c +.5, so γ 1/ 1 β 2 4

3 , and γβ Thus, the Lorentz transformation from frame S to frame S is given by: E e/c p x p y p z γ γβ γβ γ (861) 3 (861) E e /c p x p y p z (8) (9) kev/c. (1) (c) (5 points) What is the kinetic energy of the accelerated electron from the proton s point of view? In any frame, the kinetic energy is the difference between the total energy in that frame and the rest energy, or in this case: K e E e m e c 2 ( ) kev 91 kev. (d) (5 points) From the proton s point of view, what is the angle θ between the +x direction and the direction of motion of the accelerated electron? In the S frame, the angle between the +x +x direction and the motion of the electron is obtained using the relationship p x p cosθ. This relationship is true in any frame (it is basically the definition of the x-component of momentum). So we have: cos θ p x p θ cos 1 ( 96 (96) 2 + (693) 2 ) (11) y' e - θ Frame S' x'

4 4 Problem 3 Blackbody Radiation The filament of a light bulb, with length l 2 mm and radius r.5 mm, is maintained at a temperature T by an electric current. The filament behaves approximately as a black body, emitting radiation isotropically. At night, you observe the light bulb from a distance D 1 km with the pupil of your eye fully dilated to a radius ρ 3 mm. (a) (5 points) If the total power radiated by the filament is 5 watts, what is the filament s temperature? The Stefan-Boltzman Law is: dp da σt 4, (12) where dp/da is the power radiated per unit surface area of the source, σ W m 2 K 4 is the Stefan-Boltzman constant, and T is the temperature in Kelvin. Multiplying both sides of the equation by da and integrating over the surface area of the filament, gives: P tot AσT 4 2πrlσT 4 (13) ( ) 1/4 Ptot T (14) 2πrlσ [ ] 5 W 1/4 2π( m)(.2 m)( W m 2 K 4 (15) ) 3821 K. (16) (b) (5 points) What fraction of the total radiated power enters your eye? Since the lightbulb emits radiation equally in all directions (isotropic), the fraction of power entering your eye is just the cross-sectional area of your eye s pupil divided by the total surface area of a sphere centered at the source and of radius D 1 km. Thus, we have: P eye π(3 1 3 m) 2 P tot 4π(1 4 m) (1) (c) (5 points) At what wavelength does the filament radiate the most power? Here, we simply use Wien s Law: λ max b T m K 3821 K.6 1 m 6 nm. (18) This isn t too far off our experience: a real 5 bulb probably has a peak emission wavelength that is a little shorter than the calculation above, maybe around 6 nm. (d) (5 points) How many radiated photons enter your eye every second? You can assume that the average wavelength for the radiation is λ 6 nm. The energy of a photon is given by E photon hc/λ. So, dividing the total power (energy per unit time) entering your eye by the energy of an average photon, we get: N photons P eye hc/ λ. (19)

5 5 Using Eqn. 1 above and P tot 5 W then gives: N photons ( W)(6 1 9 m) ( J s)(3 1 8 m/s) photons/s. (2) Problem 4 The Step-Down Potential A beam of particles, each with energy E > and mass m, is incident from the left (x < ) on a step-down potential of the form: {, for x (region 1), U(x) (21) U, for x > (region 2). The time-independent Schrödinger equation thus reduces to: d 2 ψ(x) dx 2 { k1 2 ψ(x), in region 1, k2 2ψ(x), in region 2, (22) where k 1 2mE/ and k 2 2m(E + U )/. The general form of the solution is ψ(x) Ae ik 1x + Be ik 1x in region 1, and ψ(x) Ce ik 2x + De ik 2x in region 2. The time dependent wavefunction is given by Ψ(x, t) ψ(x)e iet/. (a) (5 points) Sketch the potential. Why can the amplitude coefficient D be set to zero? E > U(x) x -U There is no potential barrier in the region x >, so there can be no wave coming from the right region 2. Since the solutions with negative exponential coefficients correspond to waves moving from right to left, D must be equal to zero. (b) (5 points) Use the continuity equations at x to find relationships between the remaining three amplitude coefficients. In words, the continuity principle demands that the wavefunction and its first derivative be continuous everywhere, and in particular at x. This gives us the following equations: ψ 1 (x ) ψ 2 (x ) A + B C, (23) dψ 1 dx (x ) dψ 2 dx (x ) k 1A k 1 B k 2 C, (24) where we have used the fact that e 1.

6 6 (c) (5 points) The waves with amplitude A, B, and C correspond to incident, reflected, and transmitted particles, respectively. Find B and C in terms of A, k 1, and k 2. Substitute A + B for C in the second equation from part (b) above. expression for B in terms of A: This gives an k 1 A k 1 B k 2 A + k 2 B B k 1 k 2 k 1 + k 2 A. (25) Using A+B C and substituting for B from Eq. 25 gives an expression for C in terms of A: ( C A + B C 1 + k ) 1 k 2 A 2k 1 A. (26) k 1 + k 2 k 1 + k 2 (d) (5 points) The intensity (number of particles per unit time) of the incident beam is given by I A k 1 A 2 /m. Use this and the analogous relationships for the reflected and transmitted particle beams to find the reflection R(k 1, k 2 ) and transmission T (k 1, k 2 ) coefficients for particles incident from the left. Verify that R + T 1. The reflection coefficient is given by: R I B I A k 1B 2 /m k 1 A 2 /m B2 A 2 The transmission coefficient is given by: ( ) k1 k 2 2. (2) k 1 + k 2 T I C k 2C 2 /m I A k 1 A 2 /m k 2C 2 ( ) 2 k 1 A 2 2k1 k 2 4k 1k 2 k 1 + k 2 k 1 (k 1 + k 2 ) 2. (28) Computing R + T explicitly gives: R + T ( k1 k 2 k 1 + k 2 ) 2 + 4k 1k 2 (k 1 + k 2 ) 2 (29) k2 1 + k2 2 2k 1k 2 + 4k 1 k 2 (k 1 + k 2 ) 2 (k 1 + k 2 ) 2 1. (3) (k 1 + k 2 ) 2 (e) (5 points) Draw a representation of the real (not complex) wavefunction at a particular moment in time. For simplicity, you can neglect the reflected portion in your diagram. Are there bound states? There are no bound states since there is no finite-sized region where E < U(x). Note that the wavelength decreases in the region x > since the kinetic energy, and hence momentum, is larger there and since λ h/p (DeBroglie). Also note that the amplitude decreases for x >. This follows from the fact that the amplitude squared gives the probability density and since the particles speed up in the region x >, the particles spread out in that region and thus it is less likely to find a particle at a particular location.

7 U(x) x Problem 5 The 3d Subshell of the Hydrogen Atom An electron is in the 3d-subshell of the hydrogen atom. (a) (5 points) In this subshell, how many different quantum states for the electron are allowed? Enumerate the allowed states by listing the different possible combinations of quantum numbers n, l, m l, and m s. The 3d subshell has n 3 and l 2. The value of m l can run from l to l, so in this case we can have m l, ±1 or ±2. For each value of n, l, and m l, there are two possible values of m s : ±1/2. Thus there are 2(2l + 1) 1 possible quantum states in the 3d subshell of the hydrogen atom. (b) (5 points) What are the total energy and angular momentum for the 3d electron? Please do not compute numerical values! The total energy is given by: E n 1 me 4 n 2 2(4πɛ ) 2 2 E 3 1 ( 13.6) ev, (31) 9 and the total angular momentum is given by: L l(l + 1) L 6. (32) (c) (5 points) An electron in the 3d state orbits such that the probability of finding it in the x y plane is zero. What is the value of m l for this orbital? The quantum number m l quantizes the z-component of angular momentum. In order for this to be non-zero, the electron must orbit in such a way that it spends some time in the x y plane. So, in this case, we must have m l.

8 8 Problem 6 Omega Chemistry The omega particle, Ω, has the same charge as an electron but has a mass of 162 MeV/c 2 and a spin quantum number, s 3/2. Imagine that all atoms had omega particles instead of electrons. (a) (5 points) What is the intrinsic angular momentum of an Ω particle? The intrinsic angular momentum is given by: S s(s + 1) 2 2 (33) 2 (b) (5 points) What is the ground-state energy of the first element in the new periodic table? (Assume the nucleus is much more massive than the Ω particle so that it remains stationary while the Ω orbits.) Equation 31 above gives the energy levels of the hydrogen atom, where m is the mass of the electron. With the assumptions stated in the problem (nucleus much heavier than the Ω particle - i.e., lots of neutrons in the nucleus), we just need to replace the mass of the electron with the mass of the Ω particle. The ratio of Ω mass to electron mass is: m Ω 162 MeV/c2 m e 511 kev/c 2 322, (34) so all the energy levels will be 322 times more negative than in hydrogen. If you think about this, this must mean that the Ω particles stay much closer to the nucleus since the Coulomb potential gets narrower and narrower for more negative energies. So the ground state energy of the first element in the new periodic table has energy E 1 322( 13.6) ev 44, 5 ev. The innermost Ω particle is bound very tightly to the nucleus: no chemical reactions would be possible at reasonable energies (temperatures)! (c) (5 points) In accordance with the Pauli exclusion principle, how many Ω particles can occupy a single spatial state (orbital)? Explain. For each spatial state (orbital), which corresponds to a particular combination of n, l, and m l, there are 2s possible values for m s : ±1/2, ±3/2. So each orbital can accommodate four Ω particles, each with a different value of m s. (d) (5 points) What is the atomic number of the new element with a completely full 2s subshell and an empty 2p subshell? Explain. The quantum states will fill up from lowest to highest energy, with one Ω particle per unique quantum state (unique combination of n, l, m l, and m s ). Note that since the 2s subshell is filled, the 1s subshell must also be filled. In the case of electrons with s 1/2, each orbital can accommodate two electrons, one with m s 1/2 and one with m s +1/2. However, in our new chemistry, s 3/2, so each orbital can accommodate four Ω particles as explained above. Both subshells can accommodate four Ω particles since both the 1s and 2s subshells have only 2l + 1 2() orbital. Thus to fill up both the 1s and 2s subshells requires a total of eight Ω particles, corresponding to an atomic number of 8.