Astro Stats and Measurement
Measurement Problem Sheet 4 : Solutions
Andy Lawrence, University of Edinburgh
Sept-Dec 2013
(1) Photo-electric current
Consider a very bright star, with B=8.0, measured on a 4m telescope by a device which converts the
photons into a current via the photoelectric effect. The zero point of the B-magnitude system is 4260
Jy, and a typical B-band filter is centred at 440nm with an effective width of 97nm.The telescope optics, camera optics, and detector have efficiencies of 80%, 50%, and 70% respectively. If there is no
amplification, what would be the current be in amps?
One Jy is 10−26 W m−2 Hz−1 so B=8.0 corresponds to a monochromatic flux of Sν = 4260 × 10−26 ×
10−8.0/2.5 = 2.69 × 10−26 W m−2 Hz−1 . Let us assume that Sν is constant over the width of the B-band.
The width of 97nm corresponds to ∆ν = 1.60 × 1014 Hz, so the flux over the band is S = 4.30 × 10−12
W m−2 . The average energy of a photon is E = hc/λ = 4.51 × 10−19 J, so the flux of photons is
F = 9.54 × 106 photons m−2 s−1 . The rate detected by the CCD will then be R = F × πD2 /4 × 0.8 ×
0.5 × 0.7 = 1.07 × 107 photons s−1 .
This would then be the current in electrons s−1 . One amp is one Coulomb per second, and the charge
on one electron is e = 1.602 × 10−19 C, so the resulting current would be 1.71 × 10−12 Ampere. A very
very small current, even for such a bright star !
(2) Cerenkov splash energy
A 1 TeV gamma-ray produces approximately 100 Cerenkov photons/m2 at the ground, spread over a
radius of 130m, and within a ∼ 2 nsec time window. The emitted wavelength peaks around 325nm.
What is the total energy of the photons in the splash? How does this compare with the original photon
energy? Can you think of possible reasons for the difference?
For the above Cerenkov event, during the 2 nsec flash, what is the rate of arrival of photons? How many
photons per second do you get from Vega in the V-band? How do these numbers compare?
The area of the pool of light is A=53,092 m2 . (It may not be circular, but we are only getting a rough
estimate.) So we get 5.3 × 106 photons in total. At λ = 325nm, ν = 9.22 × 1014 Hz and so each photon
has energy hν = 6.11 × 10−19 J. So the total energy in the visible light splash is 3.23 × 10−12 J, which is
∼20.2MeV.
This is a factor of 50,000 short of the original gamma-ray energy, but this probably not unreasonable.
Firstly, some of the light made is absorbed in the atmosphere on the way down. Secondly, and more
importantly, only a fraction of the energy gets turned into radiation, as the secondary particles will lose
energy by all sorts of other mechanisms in the atmosphere. It is likely that most of the energy ends up
as a small amount of heat in the atmosphere.
(3) CCD charge transfer efficiency
To read out a 64 Mpix CCD camera, how many times does the charge in each pixel have to be transferred?
If we wish to lose less than 1% of the charge, how efficient must charge transfer be?
A camera with 64Mpix, if assumed square is 8000 pixels on a side. Each column is shifted over by one
column until reaching the readout column, which therefore requires 8000 transfers. Once in the readout
column, the pixles are shifted down one at a time, requiring another 8000 transfers. So 16,000 transfers
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Astro Measurement Solns-4
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in total are required.
If the efficiency per transfer is E then after N transfers the fraction remaining is E N . So to lose 1% we
need E = 0.991/N which for N = 16000 gives E = 0.9999993718542064.
(4) X-ray counts pile-up
A quasar produces an X-ray flux of ∼ 10−3 keV cm−2 s−1 keV−1 at energy 1 keV. Suppose this is being
observed by an X-ray telescope with an effective collecting area of 500 cm2 , a pixel size of 0.4900 , and an
imaging resolution which gives an 80% power radius on axis of 0.68500 . Each data frame has a standard
integration time of 3.2s. Does the quasar have a “pile up” problem ? Quantify the probabilty of there
being a problem using the Poisson distribution.
The answer depends on the range of photon energies included, so this is approximate. In a width of 1
keV, we will have 0.5 counts/s from the quasar, so over the frame time of 3.2s we will get 1.6 counts on
average. However, this will not all fall on the same pixel. Within the 80% power radius, the number
of pixels is π(0.685/0.49)2 = 6.13 pixels so we get 0.26 counts/pixel. (Actually the count rate will be
higher towards the centre). This means it is not badly piled up, but there is a serious possibility of 2
counts in the same pixel. The number of counts will follow a Poisson distribution
f (λ|k) =
λk e−λ
k!
With expected value λ = 0.26, P(0)=0.771, P(1)=0.200, P(>1)=0.028. In other words, we have a 3%
chance of pile up.
If the count rate was three times higher, we would get an 18% chance of 2 counts or more.
(5) CCD dark current
At room temperature (290K) a Silicon CCD has a dark current of ∼ 10,000 electrons/sec/pixel. What
is the dark current when used at a typical operating temperature of 200K?
The relevant formula from section 4.3.2 is
q∝T
1.5
exp
−Eg
2kT
So with Eg = 1.2ev=1.92 × 10−19 J, T1 = 290 and T2 = 200, we get q1 /q2 = 28158, and so the dark
current becomes 0.36 electrons/pixel/second.
(6) Bolometer performance
A submm source with flux 500 mJy at a wavelength 450µm is observed with a bolometer on a 15m
diameter telescope. Assuming 100% efficiency, and doing the usual νSν approximation to broadband
flux, what is the power being absorbed by the bolometer?
Germanium has a specific heat capacity of 23.22 J/mol/K. Its molar mass is 72.64 gm, and its density
is 5.23 gm cm−3 . If the active component of the bolometer is a 10µm block of Germanium, how much
energy is needed to raise its temperature by 1 milliKelvin? How long would this take when observing the
above source?
Sν =500 mJy is 5 × 10−27 W m−2 Hz−1 . At 450µm ν = 6.66 × 1011 Hz so the broad band flux is therefore
νSν = 3.33 × 10−15 W m−2 . For a 15m diameter telescope the collecting area is 176.7 m2 , so assuming
100% efficiency, the power being collected is 5.88 × 10−13 W.
The Germanium data needs to be mangled into a more useable form. (Watch out for unit changes).
Astro Measurement Solns-4
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The heat capacity is quoted as 23.22 J mol−1 K−1 - in other words, if we have one mole of Germanium,
it would take 23.22 Joules to raise its temperature by 1K. For Germanium, one mole is 72.64gm, so we
can express the heat capacity as 320 J kg −1 K−1 .
How many kg is a 10µm block of Germanium ? The volume is 10−15 m3 and the density is 5.23 g cm−3
= 5230 kg m−3 , so the block is 5.23 × 10−12 kg. We need 320 Joules for 1 kg to be raised 1K, so for
this block to be raised 1 milliKelvin we need 1.67 × 10−12 J.
So at the heating rate we calculated above for the astronomical source, 5.88 × 10−13 W, it would need
2.85 seconds to raise the temperature by one milliKelvin.
(7) Dipole size
How big is a dipole optimised to operate around the wavelength of the neutral Hydrogen (HI) fine structure
line?
A dipole of length l has a resonant frequency at ν = 4c/l. The neutral hydrogen line is at wavelength
21.106cm, corresponding to frequency 1420.406 MHz. The required value is therefore l = 8.4 cm.
(8) Heterodyne receiver
A simple heterodyne receiver operates at a fixed IF of 30MHz. We are observing HI in a galaxy with
with recession velocity 7795 km s−1 . What frequency should we tune our local oscillator to?
The intermediate frequency νIF = 30MHz is the beat frequency between the local oscillator frequency
νLO and the observed frequency ν. The rest frequency for the HI lines is 1420.406 MHz, but it is Doppler
shifted by factor 1 + z where z = v/c and v = 7795 km s−1 i.e. z = 0.026, so that ν = 1384.411MHz.
So νLO = νIF + ν = 1414.411MHz.
(9) Receiver noise
In absence of incoming radio power, the voltage V in a radio receiver will fluctuate around zero, depending on the size of the thermal noise. In the presence of incoming power, the thermal
noise will
√
¯
2
produce fluctuations around some positive value. However, the output of a receiver is V rather than
V̄ . By considering a signal with a mean of say 10 units and and noise of say 2 units and considering
just three values with man and plus and minus 1σ, show directly that noise adds a bias to the output of
the receiver.
p
If we take three values 8,10, and 12 then their mean is 10, but the RMS is (82 + 102 + 122 )/3 = 10.13.
The RMS is always systematically larger than the mean. Expressed in equivalent temperature terms,
the temperatures add.