Leaving Cert Physics Long Questions 2017 2002 15. Particle Physics
Calculate the kinetic energy of one of the particles produced, each of which has a rest mass of 9.1 × 10
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Calculate the kinetic energy of one of the particles produced, each of which has a rest mass of 9.1 × 10–31 kg. Energy of incident photon = energy required to create 2 particles + kinetic energy of particles Energy of incident photon = hf E = (6.6 × 10-34)( 3.6 × 1020) = 2.376 × 10-13 J Energy required to produce the two particles = 2[mc2] E = 2(9.1 × 10-31)(3.0 × 108)2 = 1.638 × 10-13 J Energy of incident photon = energy required to create 2 particles + kinetic energy of particles 2.376 × 10-13 J = 1.638 × 10-13 J + kinetic energy Extra energy available for kinetic energy = (2.376 × 10-13) – (1.638 × 10-13) = 7.38 × 10-14 Kinetic energy per particle is half of this = 3.69 × 10-14 Joules
Deduce the charge of each combination and identify each combination.
James Joyce 2009 Question 10 (a)
They were accelerated by the very large potential difference which existed between the top and the bottom
They collide with a zinc sulphide screen, where they cause a flash and get detected by microscopes.
 +   + K.E.
Loss in mass: Mass before = mass of proton + mass of lithium nucleus = (1.6726 × 10–27) + (1.1646 × 10–26) = 1.33186 × 10-26 kg Mass after = mass of two alpha particles = 2 × (6.6447 × 10–27) = 1.32894 × 10-26 kg Loss in mass = (1.33186 × 10-26) – (1.32894 × 10-26) = 2.92 × 10-29 kg E = mc2 = (2.92 × 10-29) (2.9979 × 108)2 = 2.6 × 10-12 J
The atom is mostly empty space so the protons passed straight through.
When the protons collide into each other they lose their kinetic energy and it is this energy which gets converted into mass to form the new particles.
The maximum that can be created would occur if all of the kinetic energy was converted into mass. Total energy = 4 GeV G = Giga = × 109 1 eV = 1.6 × 10-19 Joules 4 GeV = (4 × 109) (1.6 × 10-19) = 6.4× 10-10 Joules
You can achieve greater (particle) speeds with a circular accelerator / They take up less space 2008 Question 10 (a)
Strong (short range), Weak (short range), Gravitational (infinite range), Electromagnetic (infinite range).
Up, top, charm.
Baryon: three quarks Meson: one quark and one antiquark
Up, up, down
The total charge beforehand was +2 (due to the two protons). Therefore the total charge afterwards must be +2 (due to conservation of charge). But there are also 2 protons afterwards, so they account for the +2 by themselves. So the net charge (or all other particles added together) must therefore be zero.
P+ + P+ + kinetic energy (of protons) → P+ + P+ + π + π + π + kinetic energy So in effect the kinetic energy beforehand went into producing 3 pions, and whatever was left over became the kinetic energy of those pions plus protons. So to find this kinetic energy we need to subtract the energy required to make the 3 pions away from the original kinetic energy. Kinetic energy beforehand = 2 GeV = (2 × 109) (1.6 × 10-19) = 3.2 × 10-10 Joules Energy required to produce 3 pions: E = 3mc2 = 3(2.4842 × 10–28)( 2.9979 × 108)2 = 6.6981 × 10–11 Joules Kinetic energy after collision = (3.2 × 10-10) - (6.6981 × 10–11) = 2.53 × 10–10 J
Kinetic energy beforehand = 3.2 × 10-10 Joules Energy required to produce 1 pion = mc2 = (2.4842 × 10–28)( 2.9979 × 108)2 = 2.2327 × 10–11 Joules Number of pions =  = 14.35
So the maximum number that could have been created is 14 pions. 2007 Question 10 (a) 
See diagram.
 =  = 1.16 × 107 m s-1
 +
(accept α for He )
Mass beforehand (mass of reactants) = 1.1646 × 10-26 + 1.6726 × 10-27 = 1.33186 × 10-26 kg Mass afterwards (mass of products) = 2(6.6443 × 10-27) = 1.32886 × 10-26 kg Loss in mass = 1.33186 × 10-26 kg - 1.32886 × 10-26 kg = 3.00 × 10-29 kg
Both have equal mass / charges equal / charges opposite in sign
Pair annihilation occurs.
Fermi (and Pauli) realised that another particle must be responsible for the missing momentum, which they called the neutrino. 2006 Question 10 (a)
A photon is a discrete amount of electromagnetic radiation.
The equation for pair annihilation is as follows: 
To calculate the frequency we first need to establish how much mass gets ‘annihilated’ and then calculate how much energy that releases. = 2(1.673 × 10-27) = 3.346 × 10-27 kg
So energy associated with one photon = 1.5037 × 10-10 J We then use E = hf   f = 2.2694 × 1023 Hz
So that momentum is conserved.
They move in opposite directions.
Total charge before = +1-1 = 0 Total charge after = 0 since photons have zero charge
The energy of the photons is converted into matter .
π+ = up and anti-down π- = down and anti-up
Electromagnetic, strong, weak , gravitational
  T1/2 = 2.8 ×10-13 seconds
2005 Question 11 (a)
An alpha particle is identical to a helium nucleus (2 protons and 2 neutrons).
The atom was mostly empty space with a dense positively-charged core and with negatively-charged electrons in orbit at discrete levels around it.
Alpha particles and protons are charged, neutrons are not.
 + + K.E.
Circular accelerators result in progressively increasing levels of energy and occupy much less space than an equivalent linear accelerator.
The kinetic energy of the two protons gets converted into mass.
Up, down, strange, charm, top and bottom.
Up, up, down. 2004 Question 10 (a)
Strong nuclear force: acts on nucleus/protons + neutrons/hadrons/baryons/mesons, short range Gravitational force: attractive force, inverse square law/infinite range, all particles Electromagnetic force: acts on charged particles, inverse square law/infinite range
Mass before = mass of neutron = 1.6749 × 10–27 kg Mass after = mass of proton + mass of electron = 1.6726 × 10–27 + 9.1094 × 10–31 = 1.6817 × 10–27 kg Loss in mass (mass defect) = (1.6749 × 10–27 kg) – (1.6817 × 10–27 kg ) = 1.3891 × 10-30 kg
Momentum and energy are conserved when the momentum and energy of the associated neutrino are taken into account.
Fission is the splitting of a large nucleus into two smaller nuclei with the release of energy.
It slows down the fast neutrons (so that they in turn can be captured by the uranium atoms and cause the uranium nuclei to undergo fission).
They absorbed the neutrons which would otherwise cause fission. 2003 Question 10 (a)
Give (i) an example, (ii) a property, of each of these particles. LEPTONS; electron, positron, muon , tau, neutrino Not subject to strong nuclear force BARYONS; proton, neutron Subject to all forces, three quarks MESONS pi(on), kaon Subject to all forces, mass between electron and proton, quark and antiquark
The energy associated with the gamma ray photon (E = hf) needs to be equal to the energy associated with 2 electrons (E = 2mc2) hf = 2mc2 (6.6 × 10–34)(f) = 2(9.1 × 10–31)( 3.0 × 108)2 f = 2.5×1020 Hz
The electrons which were created would move off with greater speed. There may also be more particles produced.
e+ + e- → 2γ
Charge: The net charge of the electron and positron is 0, and there is no charge associated with the gamma ray photons. Momentum: The electron and positron are moving directly towards each other, so net momentum beforehand = 0, and afterwards the two photons move in opposite directions so net momentum after = 0. 2002 Question 10 (a)
Gravitational, Electromagnetic, Strong nuclear, Weak nuclear
Strong
Short range, act on nucleons, binds nucleus, strongest of all the forces
Protons are released at the top of the accelerator and get accelerated across a potential difference of 800 kVolt. These protons collide with a lithium nucleus at the bottom, and as a result two alpha particles are produced. The alpha particles move off in opposite directions at high speed. They then collide with a zinc sulphide screen, where they cause a flash and get detected by microscopes.
Mass before = [(1.6730 × 10-27) + (1.1646 × 10-26)] Mass after = [2(6.6443 × 10-27)] Mass defect = mass before – mass after Mass defect = 3.0 x 10-29 kg Using E = mc2 E = (3.0 x 10-29)( 3.00 × 108)2 E = 2.7 × 10-12 J Yüklə 242,08 Kb. Dostları ilə paylaş:
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m = 7.121 × 10-27 kg
Mass of particles beforehand = mass of proton + mass of antiproton