PHAS0011 Mock Examination 2022
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PHAS0011 Mock Examination
2022
One sample Astrophysics and Cosmology question worth 40% (24 marks out of 60) of an open-book exam which can be completed in 72 minutes (40% of 3 hours)
1. (a) Describe in the context of the Hertzprung-Russell diagram the different stages of stellar evolution
for a 1 solar mass star and a 10 solar mass star.
(b) How can the age of a star cluster can be measured using the distribution of stars on such a diagrams? What is the physical basis of the method?
(c) Briefly indicate during which portion of the Herzsprung-Russell diagram stars are burning helium into carbon. The initial nuclear reaction in this chain is
He4 + He4 Þ Be8 + g
If Be8 has a mass of 8.0053051 a.m.u. (or u), calculate the energetics of this reaction in eV. How does this differ from other nuclear fusion reactions that occur in stars? What conditions would be needed
therefore to initiate the reaction? [7]
(d) The black hole at the centre of our Galaxy has a mass of 4 ´106 solar masses. Calculate its Schwarzschild radius in solar units and, assuming spherical geometry, determine the average density
within the Schwarzschild radius. [4]
(e) Astronomers determined this mass by monitoring the orbits of stars in the vicinity of the black hole over many years. Assuming these orbits are circular and the distance to the Galactic centre is 8.5 kpc, what is the angular diameter of an orbit whose period is 10 years? How can such precise positional
measurements be achieved? [4]
One sample Medical Imaging question worth 20% ( 12 marks out of 60) of open-book exam which can be completed in 36 minutes (20% of 3 hours).
Using suitable diagrams, briefly describe how the molecular structure of a piezoelectric
crystal enables it to be employed to i) generate and ii) detect ultrasound pulses. [4]
A doctor working in a hospital Accident & Emergency department is responsible for deciding whether the following patients should be sent to either the MRI unit, the x-ray CT unit, or the ultrasound department for further examination:
i) A two-month-old baby who may have swallowed a wedding ring.
ii) An elderly man who is suspected of have suffered a partial blockage of blood vessels in the brain.
iii) A young woman who has splinters of glass embedded in her abdomen.
Giving reasons for your choice, state which diagnostic procedure is most appropriate in each
case, and why the other options may not be so helpful. [6]
Calculate the Doppler shift frequency obtained when a 3 MHz ultrasound probe is employed to investigate a vessel with a mean blood flow velocity of 50 cm/s, orientated at 60° to the
beam direction. The speed of sound in tissue is 1540 m/s. [2]
One sample ofModern Physics worth 40% (24 marks out of 60) of open-book exam which can be completed in 72 minutes (40% of 3 hours).
Question 3
Solutions
1 (a) [5 marks]
A descriptive account with a figure like the attached is required. Key points to make would include (i) a brief account of pre-main sequence (MS) collapse (ii) burning CNO vs pp-chain with substantially
different MS lifetimes (iii) red giant, horizontal branch (He-burning) and asymptotic giant branch phases but (iv) 1 ⊙ star does not ignite C/O/Si burning and ends as a cooling degenerate white dwarf while the 10 ⊙ star proceeds to a core-collapse supernova leaving a neutron star remnant.
(b) [4 marks]
Demonstration of physical significance of the main sequence turnoff which is age-dependent because of the temperature-dependent efficiency of nuclear fusion reactions (∝ % for pp- chain in low mass stars and ∝ '( for CNO cycle in high mass stars). Massive stars burn hydrogen much quicker so leave the main sequence first. Since all stars in a cluster can be assumed to be co-eval, there is a clear main-sequence turnoff which models can be used to age- date the turnoff point.
(c) [7 marks]
Helium ignition occurs at the uppermost tip of the red giant branch after which the star descends onto a stable He-burning `main sequence’ called the horizontal branch. The energetics of the given reaction
are:
= (2./0 − 2/3) 6 = (2 ´ 4.0026032 – 8.005305)6 = - 9.87´ 10-5 6 = - 91.9 keV
The surprising thing about this reaction is that it is endothermic, i.e. energy must be input to initiate it. A high core temperature (108 K) is thus required to generate the average kinetic energy of the gas particles c.f. hydrogen fusion.
(d) [4 marks]
The Schwarzschild radius of a black hole is 9:; = 2= 6 so for the one at the Galactic centre 9:; = (2 ´ 6.67 10- 11 ´ 2. 1030 ´ 4. 106)/ 9. 1016 = 1.18 1010 m = 17.0 solar radii
Average density = = 4. 106 ´ 2 1030 kg / ( 4.189 ´ (1.18 1010)3) m3 = 1.16 106 kg m-3 (e) [4 marks]
From Kepler’s law the period P of a star orbiting the black hole is related to the orbital radius r and the central mass M by
6 = 4F
so for a star whose period is 10 yrs, the orbital radius will be
r3 = (6.67 10- 11 ´ 4. 106 ´ 2. 1030 (10 ´ 365 ´ 24 ´ 3600)2 ´ 12.57) = 6.66 1044
So r = 8.74 1014 m (= 5825 A.U.)
At a distance of 8.5 kpc = 2.64 1020 m, this subtends an angle of only q = 3.31 10-6 radians which is
q ´ 180 ´ 3600/p = 0.68 arcsec. Tracing such an orbit requires exquisite angular resolution given typical ground-based `seeing’ is 0.5- 1.0 arcsec. This work has been accomplished using adaptive optics which corrects for the blurring in the Earth’s atmosphere using a laser-generated artificial star.
2. a)
The diagrams and the description must indicate the following:
• A piezoelectric crystal contains a regular structure of molecules (electric dipoles) aligned along a particular direction. Each molecule is physically elongated with a different electrical charge at each end. [ 1 mark]
• When the crystal is placed in an electric field the dipoles attempt to align with the field by rotating about their centres. This causes the crystal lattice to elongate in the direction of the field. [ 1 mark]
• In generation mode, the sudden application of an electric field produces a sudden distortion of the crystal. This produces a vibration = an ultrasound pulse. [ 1 mark]
• In detection mode, a mechanical compression (or elongation) of the crystal causes the dipoles to rotate. This changes the electrostatic charge on the opposite surfaces of the crystal. If the surfaces are joined in a circuit, electrons in the connecting wire will flow. [ 1 mark]
b) i) Ultrasound is the best option for detecting a ring swallowed by the infant. It is safe and would show excellent contrast between the ring and surrounding soft tissues because of the large differences in acoustic impedance. [ 1 mark]
X-ray CT would be effective, but the dose of x-rays would be particularly hazardous for a baby. [ ½ mark]
MRI would not be appropriate for two reasons: the metallic ring would produce strong image artefacts and the infant is unlikely to remain motionless during a scan lasting several minutes. [ ½ mark]
ii) Either X-ray CT or MRI would probably detect a partial block of blood vessels, although both may require a contrast agent. X-ray CT is probably more likely as it is quicker and less expensive. [ 1½ marks]
Ultrasound would not be appropriate due to absorption/reflection by the skull. [½ mark] iii) All three options could be justified. Each has potential benefits and drawbacks. [ ½ mark]
X-ray CT would reveal the glass splinters with high resolution and contrast (necessary if the splinters are very small), but the x-ray dose would be high and not appropriate if the young woman is pregnant. [½ mark]
The MRI would not show small splinters, and splinters themselves would not produce an MRI signal. Nevertheless, MRI is safe and would give 3D images. [ ½ mark]
Ultrasound would be safe, and effective if the splinters are near the surface and are not too numerous (reflections by splinters near the surface would prevent ultrasound reaching deeper splinters). [ ½ mark]
c) Doppler shift frequencyfd = ‒2 u f cosθ / v [ 1 mark] = 2 x 0.50 x 3 x 106 x 0.5 / 1540 = 974 Hz [ 1 mark]
Solution to Question 3
(a)
In the photoelectric effect, the minimum energy that is needed to eject an electron from the surface of a metal is
F
.
= ℎIJK = Φ ⇒ IJK = = = 1. 17 × 10'[
[2] points
(b)
[2] points for each of (i), (ii), [1] point for (iii). (labelled below as a,b,c]
wavelengths λ = c/f
(i) λ = 1.62 x10-27 m
(ii) λ = 1.73 x 10-35 m
(iii)λ = 4.78 x 10-26 m
(c) (i) 2 points (1 for each case)
Dt[s] |
DEmin [J] |
DEmin [eV] |
|
|
|
|
|
|
|
|
|
2 |
2.610-35 |
1.610- 16 |
c) (ii) 2 points (1 for each case)
For 1J:
For 1eV: = 1.6 10-16 s
(d) Recall Schrodinger equation:
(i)^Y^_ = a 1 − d exp(− )
[1 point]
= i − ja3 − d exp(− )
[1 point]
Substituting this with E=0 in Schrodinger equation gives
[2 points]
(ii)
[1 point]
e) (i) The normalisation condition in 1-D is
||6 = 1
Where the limits are (- infinity) to (+infinity).
In our case
6 l = 1
over the limits (a, 4a), Therefore C2 (4a-a) = 3a C2 = 1,
so C = (3a)- 1/2
[1 point]
(ii) Dimensions [C] = (length)- 1/2
[1 point]
(iii) The probability of for finding the particle in (a < x < 3a) is via the integral over these
limits
= 6 ∫ = (1/3a) (3a-a) = 2/3.
[2 points]
(f) In 3D the probability P is proportional to r2 |Y|2 = r2ne-2 r/na0
The maximum probability is at dP/dr = 0,
2 r 2n- 1 [n– r/(n a0)] e-2 r/n a0 = 0,
So the maximum probability occurs at r = n2a0 .
The radii are therefore quantised, per quantum number n, times Bohr’s radius a0
[4 points]
2022-05-17