Unformatted Document Excerpt
Coursehero >>
New York >>
SUNY Stony Brook >>
PHY 251
Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
251 PHY251_M2_S01_solutions
Name:
PHYSICS  Final Exam
Wednesday, May 16, 2001
Show all work for full credit! I. Short answer problems (120 points)
1. Calculate the momentum p and the wavelength l of: (Use eV/c and nm units!) i. a photon of 10 eV energy, Solution: p=E/c=10 eV/c; l=h/p=1240 eV/c nm/10 eV/c = 124 nm. ii. an electron of 10 eV energy, Solution: p=(2mE) = (20.51110610) = 3.2 keV/c; l=h/p=1240 eV/c nm/3.2 keV/c = 0.388 nm. iii. an electron of 10 GeV. Solution: p=E/c=10 GeV/c; l=h/p=1240 eV/c nm/10 GeV/c = 124109 nm = 0.124 fm. 2. A laser emits a beam of 1017 photons per second, of wavelength 310 nm, which falls on a piece of blackened foil (mass 20 mg). Assume the photons are fully absorbed. i. Calculate the power P of the laser beam. Solution: E = hc/l = 4.0 eV; P = NE = 10174.0 eV/s = 64 mW ii. Calculate the initial acceleration of the foil. Solution: p = E/c; a = F/m = Ndp/(mdt) = NdE/(mcdt) = P/(mc) = 64103 Nm/s /(2105 kg 3108 m/s) = 1.07105 m/s. 3. Four indistinguishable particles (with masses m = 5.0105 eV/c) are inside a cubical box, with sides equal to 1.0 nm. There is no magnetic field. i. If the particles have spin s = 0, calculate the ground state energy of the total system. Solution: The lowest energy level is at E111 = 3 /8 The next level is E211=E121=E112 = 6 /8
h mL h mL
= 30.384 eV = 1.15 eV.
= 60.384 eV = 2.30 eV.
In the ground state, all four bosons are in the first level: Etot = 4E111 = 4.61 eV. ii. If the particles have spin s = 1/2, calculate the ground state energy of the total system, and list the possible spin orientations of the particles. Solution: Only two fermions may occupy the lowest energy level with opposite spins. The other two, with opposing spins, must go into the first excited state; thus: Etot = 21.15 eV + 22.30 eV = 6.91 eV. iii. If the particles have spin s = 3/2, calculate the ground state energy of the total system, and list the possible spin orientations of the particles. Solution: Now all fermions can go into the ground state, differentiated by their ms quantum number: ms = +3/2, +1/2, 1/2, 3/2. The total ground state energy is as in (i): 4.61 eV. 4. Compton scattering of 12.4 keV Xrays: i. Under what condition is Compton scattering observed? Solution: It assumes that the scattering is off quasifree electrons! ii. Calculate the wavelength of Xrays Comptonscattered at a 60 degree angle with respect to the incident Xrays. Solution: l = 1240 eVnm /12,400 eV = 0.1 nm; l' = l  h/mc(1  1/2) = 0.100 + 1240/0.511106 = 0.1012 nm.
file:///C/Documents%20and%20Settings/Linda%20Grau...20HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (1 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
5. Xrays bombarding heavy atoms can be used to eject electrons from the 1s shell in atoms; indeed, this is the starting point for Moseley's experiments. Estimate the maximum wavelength of photons required to eject an electron from the 1s shell of copper, for which Z=29. Solution: Eg = E1(Cu) = (
Zeff
/1)(13.6 eV) @ 28 13.6 eV = 10.7 keV.
Thus: l = 1240 eVnm / 10.7 keV = 0.116 nm. 6. Describe the H2+ molecular bonding (i.e. pep bonding): i. Sketch both the electron wavefunction and the probability distribution as function of the coordinate along the pp axis that leads to stable bonding (the bonding state). Solution:  ** /  \ /*  *\_ _/* __/** ____/ **    **\_ relative plus  sign  ** /  \ _/*  *\ _/**    *\_ **\__ ** \____ **** probability ____ wavefunction
**\___/** *****
_*/__*__*_______________________________*__*__\*____ x> P< separation>P ii. Sketch both the electron wavefunction and the probability distribution as function of the coordinate along the pp axis that does not lead to bonding (the antibonding state). Solution:  ** /  \ /*  *\_ _/* __/** ____/ **    **\ *\_ *\ P **** probability relative minus sign ____ wavefunction
_*/__*__*_______________\___________________________ x> P \  **__*_/*
file:///C/Documents%20and%20Settings/Linda%20Grau...20HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (2 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
\* \* \*
  
**_/ *_/ */
\*  */ \  / **  7. Bonding: define/describe in short, and give an example of : i. Ionic bonding. Solution: Bonding between a very electronegative and a very electropositive atom: one atom has a loosely bound electron, which is strongly attracted by an atom that has an almostfilled shell. Rather than shared equally as in covalent bonding, the electron is mostly belonging to the electronegative partner. Coulomb attraction between the partners leads to bonding. Example: NaCl. ii. ss Covalent bonding. Solution: Sharing of outer sshell electrons between the atomic partners. The swavefunctions add up in between the atoms, leading to significant probability for the electrons to be found inbetween the atoms, leading to a net bonding effect. Example: H2 (see also previous problem). iii. (sp) hybridization (describe the process that leads to an electronic (sp) hybrid state). Solution: Up to two s and six p electrons are possible. If an element has 4 outershell electrons, like C or Si, then typically one will go into the s subshell, and three into the p subshell. The s electron will have a spherically symmetric wave functions, whereas the p electrons will have elongated twolobed distributions. Now, because electrons are all negatively charged, they repel each other in multielectron atoms. Therefore, and because of the Pauli exclusion principle, the p electrons will orient themselves in mutually perpendicular directions: we call them px, py and pz electrons. Often a process called sp hybridization occurs in this case (4 outershell electrons); it is based on the fact that any linear combination of wavefunctions is again a valid solution of the atomic Schrdinger equation. The lone s electron can be combined with various signs with the three p electrons, to give four independent and similar (sp) hybrid solutions. These solutions look again elongated and twolobed, but with the crucial difference that one of the lobes is much more pronounced than the other, giving a strong directionality to the distribution: the sp electron is mostly on one side of the atom, in a specific direction compared to the other three sp electrons. It is like a person with four hands, all reaching out in different directions. Mutual repulsion between the electrons makes this a lower energy configuration than the original s plus three p states. 8. Consider the NaCl molecule, in which the atomic weights of Na and Cl are 27 and 37, respectively. The internuclear separation is 0.236 nm, and the "spring constant" for vibrations is k = w2 = 300 eV/nm. i. Calculate the energy difference between adjacent vibrational levels. Solution: = [2737/(27+37)] u = 15.6931.5 MeV/c = 14.54 GeV/c DE = hw = h(k/) = hc(k/ ) = 197 eVnm (300 eV/nm/14.5109 eV) = 0.0282 eV.
c
ii. Calculate the rotational energy level for angular momentum quantum number L=1. Solution: Erot(L) =  /2 = L(L+1)( /2
I L h R
) = L(L+1)(
hc /
2cR)
=
file:///C/Documents%20and%20Settings/Linda%20Grau...20HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (3 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
= L(L+1)((197 eVnm)/214.5109 eV (0.236 nm)) = = L(L+1)((197 eVnm)/214.5109 eV (0.236 nm)) = L(L+1) 2.40105 eV. For L=1: Erot = 4.81105 eV. 9. At room temperature (300 K), calculate the probability that an electronic state with energy E = 5.25 eV in a material with Fermi energy EF = 5.00 eV is occupied. Solution: FFD = (e
EE
F/
kT
+ 1)
1
= (e0.25 eV/0.025 eV + 1)
1
= 1/(e10 + 1) e
10
= 4.5105.
10. Consider a freeelectron gas at a temperature T such that kT<< EF. Calculate an expression for the electron number density N/V for electrons that have an energy in excess of EF. Show, by making the change of variables (EEF)/kT = x, that the electron number density is proportional to T. Solution: ne(E>EF) = (Ne/V) = = (8p2 me3/2/h) F
E EF
ne(E)dE =
1
EF
g(E)dE FFD(E;T) =
EF
E dE (ex + 1)
(8p2 me3/2/h)EF
kT dx (ex + 1)
1
=
= (8p2 me3/2/h)kT EF ln2 = [2 me3/2/(ph)]kT EF ln2 = (6.85 nm3) kT EF ln2. 11. Sketch a typical energy band structure for a the following types of crystalline solids. Mark the position of the Fermi level in the graphs, and clearly label the valence and conduction bands. Give the typical size of the bandgap (in eV), and name an example of each type of solid. i. Insulator, ii. Conductor, iii. Semiconductor. Solution:  cond.  band  ______ (few eV) EF ****** ______    cond.  band       cond.  band    ^   E
______
******  val.   band  ______  
****** (0.20.6 eV)  
 val.   band  SemiConductor (Si, Ge)
 val.   band  Insulator
Conductor
(DiamondC) (metalsCu)
file:///C/Documents%20and%20Settings/Linda%20Grau...20HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (4 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
12. A current of 3.0 A flows in an aluminum wire 2.0 mm in diameter. Use the table at the end for your calculations: i. Show that the electron density in Aluminum is 181028 m 3. Solution: ne = (#electrons/atom)NA[#atoms/mole]rD[g/m]/M[g/mole] = ii. Calculate the drift speed of the conduction electrons. Solution: j = I/A = enevd vd = I/Aene = = 3 C/s / (p10 6 m 1.610 19 C 181028 m 3) = 3.310 iii. Calculate the resistivity of Aluminum. Solution: r E/j = (mvd/ ) / (enevd) = /
et m enet    5 
m/s.
=
3
= 0.511106 (eV/c) / (e 1.61019 C 181028 m
0.801014 s) =

= 5.681012 V / (1.61019 A 181028 m1 0.801014 ) = 2.4510 8 m 13. A onedimensional quantum well of width 12 nm is created in a solid, and measurements show that the energy difference between the ground state and the first excited state is 0.16 eV. What is the effective mass of the electron in the well? Solution: The energy for an electron occupying the ground state in a onedimensional box is E1 = nh/ (8mL) = h/(8mL). The energy of the first excited state is E2 = 4h/(8mL). The difference between these two states is DE=3 h/(8mL) = 0.16 eV (this difference is the same as for a 3dimensional box!). Thus, the effective electron mass is: m =3 h/(8DE L) = 3(1240 eV/cnm)/(80.16 eV 144 nm) = 2.5104 eV/c = 0.049 me (= 4.510
32
kg)!
14. Give the electronic configurations of the elements Phosphorus (P), Aluminum (Al), and Iron (Fe). Solution: P = [Ne](3s)2(3p)3 = [(1s)2(2s)2(2p)6](3s)2(3p)3; Al = [(1s)2(2s)2(2p)6](3s)2(3p)1; Fe = [Ar](4s)2 (3d)6 = [(1s)2(2s)2(2p)6(3s)2(3p)6](4s)2(3d)6 15. Doping and semiconductors: i. Describe the process of doping in silicon; how does it lead to enhanced conduction? List a typical donor material, and of a typical acceptor material. Solution: Doping is the controlled addition of "impurities" to pure silicon. The dopants are either from group 3 or group 5; i.e. they either lack a fourth p electron and are likely to bind a freeroaming electron relatively tightly (acceptor impurities like Al or Ga) thus providing a hole for conduction. Impurities that instead have five p electrons, have one left over once they covalently bind with surrounding silicon atoms, i.e. their fifth electron is only lightly bound, and is easily transferred into the conduction band (donor impurities like P or As). Silicon doped with donor impurities has electrons as the majority charge carriers (nsilicon), whereas silicon doped with acceptor impurities has holes as majority charge carriers, and is called psilicon. ii. Describe the steps that lead to formation of the depletion region in a pn junction diode. Solution: When a piece of psilicon is brough in contact with nsilicon, holes and electron charge carriers cross the junction and start to "annihilate" one another: electrons from the nsilicon get trapped in the holes on the psilicon side (leading to a net negative charge there), and viceversa , until quickly a potential is established that prevents any more electrons and holes from crossing the junction. The region depleted of charge carriers thus established is called the depletion region. iii. In what direction will the diode conduct? What is the approximate magnitude of the depletion voltage in silicon? Solution: The forward voltage that allows conduction is applied positive to the pside and negative to the nside; the voltage has to be larger than the depletion voltage (about 0.50.7 V) for an appreciable current to develop. iv. Sketch the IV characteristic curve for a pn diode (label the axes!).
file:///C/Documents%20and%20Settings/Linda%20Grau...20HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (5 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
Solution:
100 mA+  Forward Current [mA]     10 mA+      _/ _/ _/ / / /  / 
 / 
+__/+> Forward ________________________/ reverse or leakage  current (<<0.1mA)  ~0.6 V Voltage [V]
II. Consider the normalized wave function of the form y(x) = C exp{ /2 }.
a
x
(50 points)
1. Show that the normalization constant C equals (ap). Solution: The value of C is such that 1 =  y(x)2 dx = C2
exp{x/
a
}dx = C2 ap ; thus C =
(ap). 2. Show that the average of x is zero, i.e. the expectation value x = 0. Solution: x =  xy(x)2 dx =  xC2 exp{x/ }dx , which is an integral of an odd function of x,
a
over an interval symmetric with respect to x=0: therefore, the result is 0: x = 0. The result makes sense physically: the wavefunction is perfectly symmetric around x=0, and therefore the average should be zero as well.
file:///C/Documents%20and%20Settings/Linda%20Grau...20HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (6 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
3. Calculate the value of Dx, i.e. the rootmeansquare of x. Solution: First, we need to calculate x =  a. Then: Dx = RMS (x  x) = x = /2 . 4. Show that the average of px is zero, i.e. the expectation value px = 0. Hint: Use the proper definition of the expectation value, and the operator identity: (px)op = ih(/ ).
x a xy(x)2
dx = C2
x
exp{ / }dx = C2 ap. =
a
x
Solution: p =  = C2
y*(x)
[ih(/ )y(x)] dx = C2
x a x x a
exp{x/ h
2a}
[ih(/ )exp{ /2 }] dx =
x x a x
exp{x/ } 2a
[ih( / )exp{ /2 }] dx = i( / )C2
a
exp{ / } dx = 0; again because
a
x
the integrand is an odd function. The particle represented by the wavefunction is equally likely to travel left or right, and the vector average is therefore zero. 5. Calculate the value of Dpx, i.e. the rootmeansquare of px. Solution: First, we calculate the value: p = 
y*(x)
[{ih(/ )}y(x)] dx = hC2
x [(x/ a
exp{x/
2a}
[( / ) exp{ /2 }] dx =
x a
x
= hC2
h a
)  1](1/ )exp{ / }] dx =
a a x
x
= ( / )C2[(1/ )
a
exp{ / } dx  
a h a2
x
exp{x/
}dx] a
= ( / )[1/2  1] = /2 .
a a
h
h
Then: Dp = RMS (p  p) = p = /
6. Check your result for the product DxDpx against the Heisenberg uncertainty relationship and comment. Solution: Thus: DxDp = /2 . For this particular wavefunction the product DxDp reaches the minimum allowed by the Heisenberg uncertainty relationship! In many other examples  see the book  the product comes out larger. III. Consider the hydrogen atom
z f h
(60 points)
1. The angular momentum zcomponent in operator form is given by (L )op ih(/ ). Show that the wavefunction y21+1 is an eigenfunction of (L )op, and calculate the eigenvalue for this wavefunction.
z
Solution: )opy21+1 (L ih(/ )y21+1 = ih(/ )[1/
z f f
(8pa0)
( /2 0)e
a
r
r/2a
0
sinq e+if] =
= ih(+i)[1/
(8pa0)
( /2 0)e
a
r
r/2a
0
sinq e+if] = h y21+1. Thus, the function is an eigenfunction of the
Lz operator, with eigenvalue +h. 2. Calculate the probability that a 1s electron in hydrogen (see the corresponding wavefunction y100 in the formula table) is found inside the volume of the nucleus. Take the nucleus to be a sphere with radius R = 1.0 fm. Solution: P100(0rR=10
15
m) = 0
R
y
100(r)
2
rdr sinq dq df = 0
R
R10(r)Y002 rdr sinq dq df =
file:///C/Documents%20and%20Settings/Linda%20Grau...20HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (7 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
= [ 0
R
R10(r)2 rdr][
Y002sinq dq df ] = [0
15
R
R10(r)2 rdr][1].
Over the very small distance under consideration here: r R = 10
m << a0, the exponential in the Radial
function R(r) is very close to unity, and we may set it to 1 without loss of precision! Then the integral reduces to: P100(0rR=10
15
m) = 0
R
R10(r)2 rdr = 4a03 0
R
rdr= 4a03 (R/3) = 4/3(R/a0) = 4/3(210 5) 1014.

3. To map the orbits with some reasonable precision, it is necessary to localize the electrons with a precision Dx = 0.5 a0. What amount of kinetic energy (in eV) is likely to be transferred to the electron that is under observation with a probe that can localize the electrons with this precision? Solution: Dp ( /2 ) = (h/a0) = 197/0.053 eV/c = 3.7 keV/c. This is the spread in momentum; because the
Dx h
average momentum is zero (see the end of problem VI.2 above!), the average p = (Dp) = 14 (keV/c). The corresponding average kinetic energy is then: K = p/2m = 14 (keV/c) / 1.022 MeV/c = 14 eV. This is a substantial energy compared to the typical binding energy of electrons in hydrogen, and the atom will be significantly perturbed. 4. A hydrogen atom is in an excited 4d state. Show in a diagram all possible transitions that can occur and that lead to emission of photons. Solution: Note: the selection rule dictates that only transitions with Dl = 1 are allowed (to first order):
4s
4p 4d==== 4f/ / / / V /
3s
3p====/ 3d/ / / / V / V / /
2s====/2p==== / / V 1s==== V / /
5. In the atomic quantum theory it is generally not possible to specify all three components of the angular
file:///C/Documents%20and%20Settings/Linda%20Grau...20HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (8 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
momentum simultaneously. Is the l=0 angular momentum state an exception to this rule? Explain! Solution: Yes, for l=0 the magnitude of the angular momentum is zero, and hence all three components are trivially zero as well. However, this state does not violate the Heisenberg uncertainty relationship because the position and the momentum of the electron in a l=0 state are still uncertain as required by the rule DpxDx h/2. 6. An unknown magnetic field produces a set of lines for a transition from a state with l=2 to one with l=1, between which there is a maximum energy difference of 6.110 4 eV. List all allowed emission lines. Calculate the magnitude of the magnetic field. Solution: The l=2 state has a total of 2l+1 = 5 magnetic substates, and the l=1 state has three. Transitions may occur between the following levels (a total of 9 allowed transitions): l=2, m=2 l=1, m=1 DEB = BB l=2, m=1 l=1, m=1 DEB = 0 l=2, m=1 l=1, m=0 DEB = BB l=2, m=0 l=1, m=1 DEB = +BB l=2, m=0 l=1, m=0 DEB = 0 l=2, m=0 l=1, m=1 DEB = BB l=2, m=1 l=1, m=0 DEB = +BB l=2, m=1 l=1, m=1 DEB = 0 l=2, m=2 l=1, m=1 DEB = +BB The transition emission line is thus split into three (assuming the splitting of the l=2 and the l=1 levels is the same), with energies DE21, and DE21 BB; with BB = 5.79105 eV/T B = 6.110
 4 
eV. Therefore B = 6.110 4/5.79105 T = 10.5 T. 7. Positronium is a shortlived bound state in which an electron e and a positron (antielectron) e+ form a sort of atom in which the two particles circle each other around their common center of mass. Using the fact that the ground state energy of hydrogen is 13.6 eV, quickly calculate the ground state energy of positronium. (Hint: positronium is like hydrogen, but with the massive proton replaced by the positron which has a mass exactly equal to the mass of the electron.) Solution: The reduced mass of positronium is m/2; therefore E = 1/2 mc = 1/4 mc = 13.6 eV /2 =  6.8 eV.
IV. Consider a particle of mass m = 0.100 MeV/c and energy E0 = 20.0 eV incident from the left (x<0) on a potential step V(x<0) = 0, V(x>0) = V0 (not a barrier!): (50 points)
a. Calculate the momentum p1 and the corresponding wave number k of the incident particle. Solution: The incident momentum is: p1 = (2mK) = (20.10106 eV 20 eV/c) = 2000 eV/c. The wave number: k = p1/h = 2000 eV/c / 197 eVnm/c = 10.2 nm 1. b. Derive that the reflection probability is R = [
kq 
/
k+q
] for the case V0<E0.
file:///C/Documents%20and%20Settings/Linda%20Grau...20HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (9 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
Solution: Continuity at the boundary: y1(x=0) = y2(x=0) A + B = C, y1'(x=0) y1/x Thus: B/A = (
kq x=0
= y2'(x=0) ikA  ikB = iqC A  B = (q/k)C, = (q/k)(A + B).
kq
/
k+q
), and: R = [
/
k+q
]
c. For a potential step of V0 = 15 eV height, calculate the wave number q of the particle on the righthand side (x>0) of the potential step and calculate the probability for reflection. Solution: q = [2m(KV0)] / h = 1000 eV/c / 197 eVnm/c = 5.1 nm1. R = [
kq
/
k+q
] = [(1/2k)/(3/2k)] = [1/3] = 0.111
d. If instead the potential step is V0 = 25.0 eV (i.e. 5.0 eV larger than the incident particle's energy!) show that the penetration probability as function of penetration depth x (x>0) is proportional to e2 , with k = 5.1 nm1. Solution: In this case the wave number q in the region behind the step will be purely imaginary: q = [2m(KV0)] / h = i[2m(V0K)] / h = 5.1i nm1.
kx
Thus, the wave function behind the step will be completely real because iq is real: y2=Ceiqx + De
iqx
. Here,
the first exponential will be an exponentially growing function of x, and cannot remain finite as the probability interpretation of the wavefunctionsquared requires! Therefore the coefficient C must vanish. That leaves: y2=De
iqx
. = De5.1 .
x 2iqx
The barrier penetration probability P as function of x is thus: P(x>0) = y2(x) = De
= De
10.2x
.
e. For this last case, calculate the mean penetration depth inside the barrier. Solution: The mean penetration is simply the mean of the falling exponential, namely the value where it reaches 1/e down from its maximum: x = 0
Dxex/a
dx / 0
Dex/a
dx = 0
xex/a
dx / 0
ex/a
dx = a/a = a = 1/10.2 nm1 = 0.10 nm.
V. The negatively charged tau particle (t, with mass = 1800 MeV/c and mean life of 0.30 ps). (30 points) a. If the tau has a relativistic energy of 36 GeV as measured in the laboratory, calculate its mean life as observed in the laboratory. Solution: g = /
E mc
= 36,000 MeV/1,800 MeV = 20.
Thus: tlab = g t0 = 200.30 ps = 6.0 ps. b. Calculate the speed of the tau (with 36 GeV energy). Solution: The speed is b = (11/g) 1  1/800 = 0.99875, and v = 0.99875 c. c. Calculate the mean travel distance of the tau in the laboratory, and compare this to the distance it would travel in a purely Newtonian world. Solution: The mean distance traveled in the lab: d = v tlab = 0.99875 c 6.0 ps = 0.99875 0.30 mm/ps 6.0 ps = 1.8 mm. Without relativity we would have found: v = (2K/m) = (2K/m) = c (72,000/1,800) = 6.3c. And the corresponding distance in the lab would have been: d = v t0 = 6.3 0.30 mm/ps 0.30 ps = 0.57 mm only. Alternatively, if we were given the speed as 0.99875c then we would find classically: ct = c0.3 ps = 0.090 mm. d. Using the Heisenberg uncertainty relationship, calculate the uncertainty in the mass energy of the tau. Solution: Dmc hc/(2ct0) = 197 eV nm / (23.0105 nm/ps 0.30 ps) = 1.1103 eV.
file:///C/Documents%20and%20Settings/Linda%20Gra...0HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (10 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
VI. Consider a collection of NA (Avogadro's number) noninteracting identical but distinguishable atoms. (30 points) The system has a ground state energy E0 = 0, and a single excited state at energy E1 = 3.00 eV. The system is at temperature T = 34815 K. Assume the atoms obey MaxwellBoltzmann statistics, and take both the ground state and the excited state to be nondegenerate. a. Calculate the ratio of the number of atoms in the excited state to the number in the ground state. Solution: At 34815 K, the energy kT = 8.617105 34815 eV = 3.00 eV. The ratio of atoms in the excited state and those in the ground state is: NE/N0 = F = e
E E/(kT)
= e3.00/3.00 = e1.0 = 0.368 .
b. Calculate the average energy of an atom in this system. Solution: The average energy is simply: E = (N E + N00)/(N + N0) = F E/(F + 1) = (0.368/1.368) E = 0.269 E = 0.807 eV.
E E E E
c. Calculate the total energy Etot of the system in eV and in Joule. Solution: The total energy is simply the total number of atoms times the average energy: Etot = NA E = 6.021023 0.807 eV = 4.851023 eV = 7.82104 J. d. Calculate the heat capacity C dEtot/ (in J/K!) of this system.
dT
Solution: C
dE
/
dT
= / (Etot) = /dT[NA e
d d dT
E/(kT)
E /(e
E/(kT)
+ 1)] = ....
Constants and Formulae
Scale Prefixes: Constants c = 2.9979108 m/s = 3.001017 nm/s, Masses:
p
k(ilo)=103, M(ega)=106, G(iga)=109, T(era)=1012. m(illi)=103, (icro)=106, n(ano)=109, p(ico)=1012, e = 1.6021019 C
f(emto)=1015
me = 0.511 MeV/c = 9.111031 kg; m = 938 MeV/c; mn = 940 MeV/c, 1u = 931.5 MeV/c = 1.6601027 kg
Planck's constant:
h = 1240 eV/c nm = 4.1361015 eVs
h ( /2 ) = 197 eV/c nm
p
h
Boltzmann constant: k = 8.617105 eV/K,; at room temperature: T 290 K, kT 0.025 eV Coulomb's constant: kc (4pe0)1, Fine structure constant: Compton wavelength:
hc
kce = 1.440 eVnm
a (kce/ ) = 1.440 eVnm/197 eVnm = 1/137.02 lC / ) = 0.00242 nm, lC / ) = 386 fm
mc mc h h
file:///C/Documents%20and%20Settings/Linda%20Gra...0HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (11 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
Bohr radius:
a0 ( C/ ) = 0.0529 nm
Avogadros number: NA = 6.0221023 /mole Bohr magneton: Laws Special Relativity: mB eh/(2m ) = 5.788105 eV/T
e
b / ,
c
v
g [ 1/
(1  b)
]
p = gmv,
E = (pc + mc4) = gmc,
p m
K = E  mc = (g 1) mc ( = /2 Bragg scattering: 2d sinq = nl Photoelectric effect: E = hf = Kmax + W = eVstop + W Stefan's law:
0
for v<<c)
R dl = sT4,
with s = 5.67108 W/m/K4
Wien's displacement l 3 6 peakT = 2.89810 mK = 2.89810 nmK law: Planck's radiation law: R(l;T) = ( /4) (8 / 4) ( / ) [1/
l l c p hc e
hc/(lkT)  1]
Compton scattering: Dl = l'  l = [h/(mec)] (1  cosq) de Broglie wavelength: Heisenberg's uncertainty relationships: Probability: Average of f(x): l = h/p = 2p/k, with k the wavenumber DxDpx h/2, DEDt h/2
P(x) = y(x), with y properly normalized, i.e.  y(x)dx = 1 or y(r)dV = 1
f(x) y (x) f(x) y(x) dV; with dV = dx dy dz = dr rdq rsinq df
rootmeansquare of Dx (x  x) variable x: Static Schrdinger equation: ( /2 ) / y(r) + V(r) y(r) = Ey(r)
m r h
Constant potential y(x) = Aeikx + Beikx, with wave number k = (2mK)/ , m the particle mass and K the kinetic energy h V=0: (1dimensional) K=p/2
m
file:///C/Documents%20and%20Settings/Linda%20Gra...0HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (12 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
Constant box potential: V = 0 inside the box (size: LxLyLz)
y(r) = (2/Lx)sin(nxp x/Lx) (2/Ly)sin(nyp y/Ly) (2/Lz)sin(nzp z/Lz) with nx, ny, nz=1,2,3,... Enxnynz = (nx/Lx + ny/Ly + nz/Lz)( /8 ), with m the mass of the particle in the box
m h
V = outside the box Harmonic potential E = (n+) hw = (n+) h(k/ ) = (n+) hn ; n = 0,1,2,... n m V = kr: with the reduced mass m =
mM
/
m+M
and m and M the masses of the particles involved.
Coulomb potential y(r) = R (r) Y (q,f) = R (r) P (q) eimlf. ; n=1,2,3,...; l=0,1,2,...,(n1); nl lm nl lm V = kcZe/r: ml =0,1,2,...,l En = ( / ) 1/2amc = 13.61 eV ( / ) , with the reduced
n n Z Z
mass m =
mM
/
m+M
, m the electron and M the nuclear mass.
Ground state and some excited states of hydrogen (Z=1): yn,l,ml(r,q,f) 1s: y100 = 1/ 2s: y200 = 1/
(pa0)
e
r/a 0 r r/2a
(8pa0)
(1  /2 0)e
a r (8pa0)
0
2p, ml=0: y210 = 1/
( /2 0)e
a r a
r/2a
0
cosq
0
2p, ml=1: y211 = 1/ 3s: y300 = 1/
(27pa0)
(8pa0) r
( /2 0)e
a
r/2a
sinq eif
r/3a
0
(1  2 /3 0+ 2r/27 0)e
a
Atomic quantum numbers:
n = 1,2,3,4,... ( K,L,M,N,... shells) l = 0,1,2,3,...,n1 ( s,p,d,f,g,... ) ml = 0,1,2,3,...,l. Transitions: Dl = 1 s = , ms =
Atomic angular momenta: Particle Spin
L = h[l(l+1)] , l=0,1,2,...,(n1); Lz = mlh, ml=0,1,2,...,l; S = h[s(s+1)] , s=; Sz = msh, ms= (for electrons, protons and neutrons) S = h[s(s+1)]; Fermions: s=1/2,3/2,5/2,..; Bosons: s=0,1,2,..; Sz = msh, ms = s,s1,s2,...0
file:///C/Documents%20and%20Settings/Linda%20Gra...0HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (13 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
Magnetic potential energy:
VB = B = zB (z//B), = ( /2 )L, with Q the charge;
me
gQ
for e: VB, = ( /2 )Lz B = mBml B;
e L me
VB, = mBgems B = mB2ms B
S
Moseley's law: Rotation:
DE = hn = Zeff(1/2mca)(3/4); Erot = [ /2 ],
I L
Zeff Z  1 L=0,1,2,... Transitions: DL = 1
mM
L = L(L+1) h ,
with the moment of inertia I = mR, with reduced mass m = Vibration: Statistics: Evib = , (N+) hw = (N+) h( / ) = (N+) hn,
m k
/
m+M
and separation distance R
N=0,1,2,...
Number of microstates: MaxwellBoltzmann:
Wj = N!/(nj,0! nj,1! nj,2! nj,3! nj,4! ... ) FMB(E;T) = AeE/kT F(v;T) = 4pn ( /2
(EEF)/kT + 1)] e E/kT  1)] Be
Maxwell speed distribution: FermiDirac: BoseEinstein:
m
pkT
)3/2 v2e
mv/(2kT)
FFD(E;T) = [1/(
FBE(E;T) = [1/(
(for photons: B = 1)
N
n(E)dE = N(E)/V dE = g(E)dE Fxx(E;T), with normalization condition: 0 n(E)dE = /V Density of states for a cubic box: g(E) = [8p2 me3/2 /h] E = [2 me3/2 /(ph)]E. Solid State Drift speed of electrons: vd = eEt/m, with t is the (quantum) Drude relaxation time; l = t vF = 1/(nis) is the mean free path, E the electric field, vF the Fermi velocity (i.e the speed of electrons with energy EF), s the cross section for electron scattering, and ni the number density of ions in the solid. Electric current density: j I/A = enevd , where A is the cross sectional area, and ne the number density of conduction electrons. Resistivity: r E/j Metal Li Al Cu Nuclear physics: Nuclear decay: Atomic weight Drude relaxation Fermi velocity Mass density Conduction [g/mole] time t [s] [m/s] [kg/m] electrons/atom 6.9 27.0 63.5 (
dN(t) dt
0.881014 0.801014 2.71014 / ) = lN(t),
1.29106 2.03106 1.57106 N(t) = N0e = N0 2 /
tt lt
0.53103 2.70103 8.93103
1 3 1
Math formulae:
0
0
xn ex/a dx
= n! an+1 ;

ex/a
dx = ap ;

x ex/a
dx = ( /2)p ; (x 1)
a
(ex + 1)
1
dx = ln 2;
ln(1+x) = x  [(x2)/2] + [(x3)/3]  [(x4)/4] +...
file:///C/Documents%20and%20Settings/Linda%20Gra...0HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (14 of 15) [2/4/2008 4:41:09 PM]
PHY251_M2_S01_solutions
file:///C/Documents%20and%20Settings/Linda%20Gra...0HEPH2/PHY251%20Sp01/PHY251_F_S01_solutions.html (15 of 15) [2/4/2008 4:41:09 PM]
Find millions of documents on Course Hero  Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.
Below is a small sample set of documents:
SUNY Stony Brook  PHY  251
PHY251 Midterm I ExamPHY 251 Midterm I ExamI. Short Answer QuestionsMarch 7 2001(110 points)1. List the crucial experimental evidence that showed that classical Newtonian mechanics was a lowvelocity approximation of relativistic mechanics. S
SUNY Stony Brook  PHY  251
PHY251_M2_S01_solutionsName:PHYSICS 251  Midterm IIWednesday, April 18, 2001Show all work for full credit! 1. Short answer problems (100 points) a. An unstable system emits a photon of l = 121.50 nm which has an uncertainty in its wavelength
SUNY Stony Brook  PHY  251
phy251_s01_hw01PHY251 Homework Set 1Reading: Chapters 1 and 2 Homework: Problems are listed belowProblems, Hints, and SolutionsProblem I.1 An atom of mass m is moving along the xdirection with speed v0. It collides elastically with an atom of
SUNY Stony Brook  PHY  251
phy251_hw02PHY251 Homework Set 2Reading: Chapter 2, 3 Homework: Chapter 2, Questions 6, Problems 2,4,7,8 Chapter 3, Questions 7, Problems 3,4,34Hints and SolutionsQuestion II.6 You have been given asuperwatch accurate to a nanosecond. You want
SUNY Stony Brook  PHY  251
phy251_hw03PHY251 Homework Set 3Reading: Chapter 4 Homework: Chapter 4, Questions 3,5,7 Problems 2,8,13,16,32Hints and SolutionsNote: SHOW ALL WORK! Question IV.3 How would you estimate the temperature of a redhot piece of steel, given that th
SUNY Stony Brook  PHY  251
phy251_hw04PHY251 Homework Set 4Reading: Chapter 5 Homework: Chapter 5, Questions 1,3,4 Problems 6,8,9,16,18Hints and SolutionsQuestion V.1 Hints: No hints Solution: No Solution yetfile:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_hw04.HTML [2/20/2001
SUNY Stony Brook  PHY  251
phy251_hw05PHY251 Homework Set 5Reading: Chapter 4 and 5 Homework: Chapter 4, Questions 2,4 Problems 4,5,7 Chapter 5, Problems 2,22Hints and SolutionsQuestion IV.2 If you peer through a small hole in a very hot oven, you see a glow that clearly
SUNY Stony Brook  PHY  251
phy251_hw06PHY251 Homework Set 6Reading: Chapter 6 and 7 Homework: Chapter 6, Question 9 Problems 2, 9,18 Chapter 7, Problems 16,20Hints and SolutionsQuestion VI.9 Suppose you have a box a few centimeters wide and an electron with the energy of
SUNY Stony Brook  PHY  251
phy251_hw07PHY251 Homework Set 7Reading: Chapter 8 Homework: Chapter 8, Question 1,4 Problems 1,6Hints and SolutionsQuestion VIII.1 In the wavefunction that describes a particle colliding with a barrier in one dimension we have elements that de
SUNY Stony Brook  PHY  251
phy251_hw08PHY251 Homework Set 8Reading: Chapter 9 Homework: Chapter 9, Question 2, Problems 11,14,15,19; (extra credit: 23)Hints and SolutionsQuestion IX.2 (10 points) In the quantum theory it is generally not possible to specify all three com
SUNY Stony Brook  PHY  251
phy251_hw09PHY251 Homework Set 9Reading: Chapter 10, with exception of Sections 105 and 106 Homework: Chapter 10, Question 4, Problems 1,3,5,6,7,11Hints and SolutionsQuestion X.4 (10 points) According to the exclusion principle, in any system
SUNY Stony Brook  PHY  251
phy251_hw10PHY251 Homework Set 10Reading: Chapter 11, 12 Homework: Chapter 11, Question 1, Problems 2,5,11,14 Chapter 12, Problems 7,23,31Hints and SolutionsQuestion XI.1 (10 points) One of the great puzzles in heredity before the quantum theor
SUNY Stony Brook  PHY  251
phy251_hw11PHY251 Homework Set 11Reading: Chapter 12, 14 (all sections except 6) Homework: Chapter 12, Problems 32, Chapter 14, Problems 2, 6(Li, Al, Cu only!),9(Li, Al, Cu only!),10,17,18,31Hints and SolutionsProblem XII.32 (15 points) Conside
SUNY Stony Brook  PHY  251
phy251_s01_hw01PHY251 Homework Set 1Reading: Chapters 1 and 2 Homework: Problems are listed belowProblems, Hints, and SolutionsProblem I.1 An atom of mass m is moving along the xdirection with speed v0. It collides elastically with an atom of
SUNY Stony Brook  PHY  251
PHY251 Homework Set 2Reading : Chapter 2, 3 Homework:{ {Chapter 2, Questions 6, Problems 2, 4, 7, 8 Chapter 3, Questions 7, Problems 3, 4, 34Question II.6 You have been given a superwatch accurate to a nanosecond. You want to set the watch so t
SUNY Stony Brook  PHY  251
phy251_hw03PHY251 Homework Set 3Reading: Chapter 4 Homework: Chapter 4, Questions 3,5,7 Problems 2,8,13,16,32Hints and SolutionsNote: SHOW ALL WORK! Question IV.3 How would you estimate the temperature of a redhot piece of steel, given that th
SUNY Stony Brook  PHY  251
phy251_hw04PHY251 Homework Set 4Reading: Chapter 5 Homework: Chapter 5, Questions 1,3,4 Problems 6,8,9,16,18Hints and SolutionsQuestion V.1 In studying the Bohr model in this chapter, we ignored special relativity. Why, then, does the speed of
SUNY Stony Brook  PHY  251
phy251_hw05PHY251 Homework Set 5Reading: Chapter 4 and 5 Homework: Chapter 4, Questions 2,4 Problems 4,5,6,7 Chapter 5, Problems 2,22Hints and SolutionsQuestion IV.2 If you peer through a small hole in a very hot oven, you see a glow that clear
SUNY Stony Brook  PHY  251
phy251_hw06PHY251 Homework Set 6Reading: Chapter 6 and 7 Homework: Chapter 6, Question 9 Problems 2, 9,18 Chapter 7, Problems 16,20Hints and SolutionsQuestion VI.9 Suppose you have a box a few centimeters wide and an electron with the energy of
SUNY Stony Brook  PHY  251
phy251_hw07_solutionsPHY251 Homework Set 7Reading: Chapter 8 Homework: Chapter 8, Question 1,4 Problems 1,6Hints and SolutionsQuestion VIII.1 (10 points) In the wavefunction that describes a particle colliding with a barrier in one dimension we
SUNY Stony Brook  PHY  251
phy251_hw08_solutionsPHY251 Homework Set 8Reading: Chapter 9 Homework: Chapter 9, Question 2, Problems 11,14,15,19; (extra credit: 23)Hints and SolutionsQuestion IX.2 (10 points) In the quantum theory it is generally not possible to specify all
SUNY Stony Brook  PHY  251
phy251_hw09_solutionsPHY251 Homework Set 9Reading: Chapter 10, with exception of Sections 105 and 106 Homework: Chapter 10, Question 4, Problems 1,3,5,6,7,11Hints and SolutionsQuestion X.4 (10 points) According to the exclusion principle, in
SUNY Stony Brook  PHY  251
sPHY251 Homework Set 10Reading: Chapter 11, 12 Homework: Chapter 11, Question 1, Problems 2,5,11,14 Chapter 12, Problems 7,23,31Hints and SolutionsQuestion XI.1 (10 points) One of the great puzzles in heredity before the quantum theory of molec
SUNY Stony Brook  PHY  251
phy251_hw11PHY251 Homework Set 11Reading: Chapter 12, 14 (all sections except 6) Homework: Chapter 12, Problems 32, Chapter 14, Problems 2, 6(Li, Al, Cu only!),9(Li, Al, Cu only!),10,17,18,31Hints and SolutionsProblem XII.32 (15 points) Conside
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 1PHY251 Lecture Notes  Set 1file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_01.HTML (1 of 15) [1/23/2001 5:19:56 PM]PHY251 Lecture Notes Set 1file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_01.HTML (2 of 15) [1/23/
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 2PHY251 Lecture Notes  Set 2file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_02.HTML (1 of 6) [2/8/2001 2:01:45 PM]PHY251 Lecture Notes Set 2file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_02.HTML (2 of 6) [2/8/2001
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 3PHY251 Lecture Notes  Set 3file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_03.HTML (1 of 9) [2/8/2001 1:58:26 PM]PHY251 Lecture Notes Set 3file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_03.HTML (2 of 9) [2/8/2001
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 4PHY251 Lecture Notes  Set 4file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_04.HTML (1 of 17) [2/20/2001 7:35:36 PM]PHY251 Lecture Notes Set 4file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_04.HTML (2 of 17) [2/20/
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 5PHY251 Lecture Notes  Set 5file:/C/Documents/phy_courses/PHY251/PHY251_lecture_05.HTML (1 of 5) [3/7/2001 2:41:57 PM]PHY251 Lecture Notes Set 5file:/C/Documents/phy_courses/PHY251/PHY251_lecture_05.HTML (2 of 5) [
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 6PHY251 Lecture Notes  Set 6file:/C/Documents/phy_courses/PHY251/PHY251_lecture_06.HTML (1 of 7) [3/7/2001 2:46:00 PM]PHY251 Lecture Notes Set 6file:/C/Documents/phy_courses/PHY251/PHY251_lecture_06.HTML (2 of 7) [
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 7PHY251 Lecture Notes  Set 7file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_07.HTML (1 of 14) [4/3/2001 12:03:18 PM]PHY251 Lecture Notes Set 7file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_07.HTML (2 of 14) [4/3/2
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 8PHY251 Lecture Notes  Set 8file:/C/Documents/phy_courses/PHY251/PHY251_lecture_08.HTML (1 of 6) [4/3/2001 12:30:05 PM]PHY251 Lecture Notes Set 8file:/C/Documents/phy_courses/PHY251/PHY251_lecture_08.HTML (2 of 6)
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 9PHY251 Lecture Notes  Set 9file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_09.HTML (1 of 14) [4/16/2001 1:51:03 PM]PHY251 Lecture Notes Set 9file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_09.HTML (2 of 14) [4/16/
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 10PHY251 Lecture Notes  Set 10file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_10.HTML (1 of 8) [4/23/2001 2:12:37 PM]PHY251 Lecture Notes Set 10file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_10.HTML (2 of 8) [4/23
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 11PHY251 Lecture Notes  Set 11file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_11.HTML (1 of 20) [4/23/2001 1:56:21 PM]PHY251 Lecture Notes Set 11file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_11.HTML (2 of 20) [4/
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 12PHY251 Lecture Notes  Set 12file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_12.HTML (1 of 13) [5/1/2001 7:44:45 PM]PHY251 Lecture Notes Set 12file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_12.HTML (2 of 13) [5/1
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 13PHY251 Lecture Notes  Set 13file:/C/Documents/phy_courses/PHY251/PHY251_lecture_13.HTML (1 of 12) [5/8/2001 9:05:47 AM]PHY251 Lecture Notes Set 13file:/C/Documents/phy_courses/PHY251/PHY251_lecture_13.HTML (2 of
SUNY Stony Brook  PHY  251
PHY251 Lecture Notes Set 14PHY251 Lecture Notes  Set 14file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_14.HTML (1 of 7) [5/8/2001 9:19:33 AM]PHY251 Lecture Notes Set 14file:/C/DOCUME~1/PHY_CO~1/PHY251/PHY251_lecture_14.HTML (2 of 7) [5/8/2
SUNY Stony Brook  PHY  251
PHY 251Modern PhysicsFinal grades available (see Grading below)Fall 1998PHY251 is a wideranging introduction to the physics of the 20th century, which saw a radical evolution in the understanding and view of the submicroscopic domain and in
SUNY Stony Brook  PHY  251
file:/C/Documents%20and%20Settings/Linda%20Grauer/My%20Doc.20HEPH2/PHY251%20Sp01/Phy251%20Fa98/PHY251_M1_F98solutn.htmlName:PHYSICS 251  Midterm IWednesday Oct 14, 1998I. Short answer problems: (60 points) a. List and/or name: (16 points)
SUNY Stony Brook  PHY  251
file:/C/Documents%20and%20Settings/Linda%20Grauer/My%20Doc.20HEPH2/PHY251%20Sp01/Phy251%20Fa98/PHY251_M2_F98solutn.htmlPHYSICS 251  Midterm IIMonday, November 23, 1998 Show all work for full credit!Short answer problems, each about 3 points:
SUNY Stony Brook  PHY  251
PHY251  MichelsonMorleyPHY 251IntroductionMichelson InterferometerTo better understand the interferometer used by Michelson and Morley, we use a simplified version of their instrument to measure the wavelength of the laser light source used.
SUNY Stony Brook  PHY  251
PHY251  Photoelectric EffectPHY 251IntroductionThe Photoelectric EffectIn this experiment you will use the photoelectric effect to measure the Planck constant h. This classical experiment led to the first precise determination of h, and in 19
SUNY Stony Brook  PHY  251
PHY251  Measurement of Electron ChargePHY 251IntroductionMeasurement of the Electron ChargeThe purpose of this experiment is to measure the smallest unit into which electric charge can be divided, that is, the charge of an electron e. The met
SUNY Stony Brook  PHY  251
PHY251  Waves on a stringPHY 251 Vibration Spectrum of 1Dimensional WavesIntroductionIn this experiment we study standing waves on a long rubber band stretched between fixed supports. These standing waves are analogous to the wave functions of
SUNY Stony Brook  PHY  251
PHY251  e/m for electronsPHY 251IntroductionMeasurement of e/m for ElectronsIn this experiment you will measure e/m, the ratio of charge e to mass m of the electron. The setup consists of an evacuated (except for a low pressure of He gas) ca
SUNY Stony Brook  PHY  251
PHY251  ScatteringPHY 251IntroductionScatteringThe purpose of this laboratory is to show how certain characteristics of an object can be determined from the distribution of particles (or radiation) scattered from the object. In particular, we
SUNY Stony Brook  PHY  251
PHY251  Hydrogen SpectrumPHY 251IntroductionEmission Spectrum of HydrogenIn this experiment you will measure the visible part of the hydrogen spectrum, the Balmer series, and determine the Rydberg constant Ry. If parallel rays of light are in
SUNY Stony Brook  PHY  251
PHY251  Bragg ScatteringPHY 251PurposeBragg Scattering of MicrowavesIn this experiment, we will study the scattering of microwaves from a `crystal' made up of ball bearings embedded in styrofoam. We will find diffraction maxima that are simil
SUNY Stony Brook  PHY  251
PHY251  Geiger counterPHY 251IntroductionGeiger CounterThis experiment has two main objectives. First, we will become familiar with the characteristics of the GeigerMller (GM) tube, including the properties of threshold, starting and operati