This blog contains answers to exercises set for students. While every effort is made to ensure that the information posted is correct, mistakes may occur from time to time.
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Thursday, April 29, 2021
Wednesday, April 28, 2021
Particle Physics Questions
1. C
2. C
3. (a)
|
hadron |
baryon |
lepton |
neutron |
y |
y |
|
proton |
y |
y |
|
electron |
|
|
y |
neutrino |
|
|
y |
4 lines correct 2/2: 3 lines correct
1/2: 2 or 1 line correct 0/2 (2) 2
3. Three
from:
a. Proton
turns into a neutron and positron
b. Positron
is released
c. Up
quark turns to down quark
d. Electron
neutrino released
[3]
5. (i) up down down / udd; 1
(ii) Q B S
u (+)2/3 (+)1/3 0 u
values (1)
d –1/3 (+)1/3 0 d
values (1) 2
(iii) so
for neutron Q = 0
B = 1
S = 0 1
[4]
6. (i) leptons; 1
(ii) neutrino / muon / tau(on); 1
[2]
7. lepton: two examples: electron; (1)
positron;
(1)
neutrino;
(1) any 2 (2)
(allow
muon, tauon)
3 particles including one wrong gets 1 only
composition: fundamental (- no quark
components); (1)
forces: weak force / interaction; (1)
electron / positron - (also)
electromagnetic / electrostatic force; (1)
where found: electron - in atom, outside nucleus or in β– decay; (1)
positron (rarely)
emerging from (high mass) radioisotopes /
in β + decay / accelerating-colliding machines; (1)
neutrino -
travelling in space eg from Sun
or emitted
(with electron / positron) in beta decay; (1)
allow ONCE ‘resulting from high energy
particle collisions’ any
6
[6]
8. baryon:
two examples proton; (1)
neutron;
(1)
3 particles quoted, including one wrong gets 1/2 only
quark composition: proton uud;
(1)
neutron udd; (1)
(aware consists of 3 quarks,
unspecified, gets 1/2)
stability: proton stable inside (stable) nucleus; (1)
proton possible decay /
half life = 1032 years
when free; (1)
allow any half life
> 1030 years
neutron stable inside
(stable) nucleus; (1)
neutron half life = 10/15
minutes when free; (1) any
5
[6]
9. neutron is udd / proton is uud; (1)
quarks are: up down strange top bottom charm; (1)
either up / u has Q = (+)2/3, B = (+)1/3;
or down / d has Q = –1/3, B = (+)1/3; (1)
quarks are fundamental particles; (1)
for every quark there is an antiquark; (1)
antiquarks have opposite values of Q, B and S (compared to
quark) (1)
quarks are held together by strong force / gluons (1)
Q, B and S are conserved in (quark) reactions (1) any 2 5
[5]
10
THREE FROM:
the strong interaction ✔
has short range OR mention range (less than 5
fm) ✔
attraction up to 5 fm ✔
repulsive (any distance below 1fm) ✔
is zero/negligible beyond 5 fm ✔
only affects hadrons/ baryons and
mesons ✔
mediated by gluons/pions ✔
If wrong interaction
identified then zero marks
If refer to strong interaction correctly then
ignore any subsequent reference to other interactions
Tuesday, April 27, 2021
Latent heat
- Calculate the
energy transferred to the surroundings when 50 g of aluminium changes
phase from liquid to solid. E = mLf E = 0.050 × 398000 = 20 kJ
- Calculate the
energy required to change 2.5 kg of silver at its melting point from solid
to liquid. E = mLf E = 2.5 × 88000 = 220 kJ
- When a kettle boils a space can be
seen between the spout and the so called `cloud of steam'. If your finger
were placed in the space a much more serious scald would result than if it
were placed in the `cloud of steam'. Explain why, and criticize the
expression `cloud of steam'. Steam is a colourless
gas which fills the space, what we see is water vapour condensing in the
air as it cools.
- A 1 kW immersion heater takes 225 s
to boil away 0.1 kg of water at 100 °C. Show that the specific latent heat
of vaporization of water is about 2.1 KJkg-1
E =Pt = mLv 1000 x
225 = 0.1Lv Lv= 225000/0.1 =
2.25 kJkg-1
- Describe why the
specific latent heat of vaporisation is normally greater than the specific
latent heat of fusion for a particular substance. There is a
greater change in internal energy changing phase from liquid to gas than
from solid to liquid
- A mixture of ice and water at 0 °C
contains 0.060 kg of ice. A 750 W immersion heater takes 26.8 s to melt
all the ice. Show that the value for the specific latent heat of fusion of
ice is approximately 3.4 x 105 Jkg-1.
E =Pt = mLv 750 x 26.8
= 0.060Lv Lv= 20100/0.060 = 3.35
x 105 Jkg-1