Physics 3401, Homework quiz, Ch 3,4

For questions 1 and 2 (which may be worked independently), you are dealing with a hot potato.
Approximate this potato with a sphere of radius 4 cm. It has an initial temperature of 100°C or 373K.
 It's mass is 320 grams.

1. a. Find the initial cooling rate in watts for this potato. Assume it to be an ideal blackbody radiator.

Power = sigma*area*T^4
area=4*pi*r^2= 0.0201 m^2
Power = 5.67e-08*.0201*373^4=  22.06 watts

If the environment is at 20 C = 293K, then there will be some radiation to the potato, so if it were a perfect absorber it would take in heat at a rate:

Power = 5.67e-08*.0201*293^4= 8.4 watts

So the net cooling rate would be 13.66 watts

b. Find the wavelength of the peak of the radiation curve.

peak wavelength = .0029/373 =  7.7748E-06 m= 7774.8 nm

c. Assuming that all photons have the energy of the Wien peak, find the number of photons per second emitted.

Photon energy of Wien peak = 1240 eVnm/7775nm =    0.159 eV

Number of photons = 22.06 watts/(0.159*1.6e-19)=    8.67E+20 photons/s

2. Suppose you want the hot potato to cool to 80°C to eat it. Calculate the radiative cooling time to this
temperature. Assume that it is an ideal radiator and that the average molecular mass in the potato is 30
(a wild guess, like two carbons and 6 hydrogens on average). Assume that each molecule has 3kT/2 of
energy. Discuss any further assumptions you make.

Cooling time = (Nk/2*sigma*A)(1/Tcold^3-1/Thot^3)
320grams would by 320/30 =  10.66666667 moles = 6.42133E+24 molecules = N

time = (6.42e24*1.38e-23J/K/(2*5.67e-8*.020m^2))*(1/353^3-1/373^3) =     135. sec
Neglecting environmental transfer back to the potato

This is too short a time, because we have neglected reabsorption of heat from the environment, which will be assumed to be 293K, giving a reabsorption rate of 8.4 watts. The net cooling rate at 80 C = 353K would be 17.70 watts with no reabsorption. The integral with reabsorpton in it is a little too hard to do, so we can estimate the lengthening of the cooling by seeing that the reabsorption slows the initial cooling by 8.4/22=38%, and the final cooling rate by 8.4/17.7 = 48%. Taking the average of43% gives cooling time estimate 135s/.57=237 sec.

3. A certain short-lived charged particle has an average lifetime of 10-13 sec. Assuming that it is
emitted at a speed of 0.98c. Being a charged particle, it leaves a visible track in a laboratory frame
detector medium. How long do you expect the average track to be?

Lab frame halflife dilated to 5e-13
Track length = 0.98*3e8*5e-13=  0.000147 m = 0.147 mm

4. You have erected a tower on top of Stone Mountain to do a local repetition of the muon experiment. You
have a detector at 500 m height and another identical one at ground level at the base of the mountain. The
rest frame half-life of the muon is 1.56 x 10-6 sec. Presuming that in an hour's time you detect 1000
muons at the top detector, how many do you expect to detect at the lower detector.  Assume that there is a
uniform flux of muons at speed 0.99c. Show that the result is the same as seen from the earth frame and
the muon frame.

Relativity factor 7.08

From earth frame, time = 500m/(.98*3e8m/s) =   1.7007E-06 sec
Halflife seen as dilated to 1.56e-6*7.08 =    1.10E-05 sec
Number of halflives =  0.154
Number at earth = 1000*(2^-.154)=   898.8

From muon frame,length is contracted to 500/7.08=   70.62 m
The time is then 70.6/(.98*3e8)=   2.40E-07
The number of halflives is then = 2.4e-7 /1.56e-6=   1.54E-01
so the final result is the same.

5. A space ship traveling at 0.6c is approached by another ship which from an external reference frame
is traveling at 0.8c toward the first ship.

a. What approach speed would the first ship measure for the second ship?

vapproach = (0.6 + 0.8)/(1+(0.6*0.8))=   0.9459 c

b.If the first ship is propelled by an ion drive which ejects ions at 0.7c compared to the ship, with what
 speed would an external observer see those ions traveling?

v = (0.6 -0.7)/(1-(.6*.7))=  -0.1724 c

6. Taking a rough estimate of the blackbody curve of a radiator for which the spectrum peaks in the middle
 of the visible range, it appears that about 37% of the radiated energy is in the visible.

a. What temperature would the radiating surface have to be to peak at 550nm?

T=.0029/550e-9=  5272.7 K

b. What area of radiating surface would be required to  produce 1000 watts of power in the visible range
if the surface is assumed to be an ideal radiator?

Total power radiated = 1000/.37 =2702.7 watts

A=2703/(sigma*T^4)= 6.167E-05 m^2 = 0.617 cm^2

7. a. Find the momentum of an electron which has a kinetic
 energy of 1000 eV and one which has a kinetic energy of 1000 MeV.
pc=sqrt((1000eV+511000eV)^2-511000^2) =   31984. eV
pc=sqrt((1000MeV+ .511MeV)^2-.511^2) =   1000.51MeV

b. Find the momentum of photons with the same energy.
pc=E for photon, so pc=1000eV and 1000MeV