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This is the Solved Exam of General Physics which includes Vector Quantities, Scalar Quantities, Resultant of Two Vectors, Circumference of Circle, Newton’s Second Law of Motion, Magnitude and Direction etc. Key important points are: Particle Physics, Particle Accelerators, Cockroft and Walton Experiment, Alpha Particles, Advantage of Circular Accelerators, Linear Accelerators, Nuclear Equation for Decay
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Exam Questions mass of proton = 1.6730 × 10-27^ kg; mass of electron = 9.1 × 10–31^ kg; mass of lithium nucleus = 1.1646 × 10-26^ kg; mass of α-particle = 6.6443 × 10-27^ kg; mass of neutron = 1.6749 × 10–27^ kg; charge on electron = 1.6022 × 10–19^ C; mass of pion = 2.4842 × 10–28^ kg; speed of light, c = 2.9979 × 10^8 ms-1; Planck constant = 6.626 × 10-34^ J s
Particle Accelerators, Cockroft and Walton Experiment and E = mc^2
The neutrino
Give two reasons why it is difficult to detect a neutrino.
Pair Production
Pair Annihilation
Fundamental Forces
Quark Composition and Particle Classification
A huge collection of new particles was produced using circular accelerators. The quark model was proposed to put order on the new particles. List the six flavours of quark.
(i) 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. (ii) Total energy = 4 GeV E = mc^2 ⇒ m = E/ c^2 ⇒ m = (4 × 10^9 ) (1.6 × 10-19)/(2.9979 × 10^8 )^2 ⇒ m = 7.121 × 10-27^ kg
(i) Zero, because electric charge must be conserved. (ii) Energy equivalent of a pion:) E = mc^2 E = (2.4842 )( 2.9979 × 10^8 )^2 E = 2.2327 × 10–11^ J = 1.3935 × 10^8 eV For 3 pions E = 6.6980 × 10–11^ J = 4.18047 × 10^8 eV Energy after collision = (2 × 10^9 ) - (4.18047 × 10^8 ) = 1.58195 × 10^9 eV = 2.535 × 10– (^10) J
(iii)Number of pions = (1.58195 × 10^9 ) / 1.3935 × 10^8 = 11.3524 = 11 pions. Maximum number of pions = 3 + 11 = 14 pions.
(i) A photon is a discrete amount of electromagnetic radiation. (ii) m [= mass of proton + mass of antiproton ] = 2(1.673 × 10-27) = 3.346 × 10-27^ kg E = mc^2 = (3.346 × 10-27^ )(2.998 × 10^8 )^2 = 3.0074 × 10- Energy for one photon = 1.5037 × 10-10^ J E = hf ⇒ f = E/h / = 1.5037 × 10-10^ / 6.626 × 10-34^ = 2.2694 × 10^23 Hz (iii)So that momentum is conserved. (iv) They move in opposite directions. (v) Total charge before = +1-1 = 0 Total charge after = 0 since photons have zero charge (vi) The energy of the photons is converted into matter.
(i) e+^ + e-^ → 2γ (ii) Total charge on both sides is zero Momentum of positron + electron = momentum of photons
[2010] Give two advantages of a circular accelerator over a linear accelerator. Smaller (less space) // greater speeds/energy
(i) What is anti-matter? Antimatter is material/matter/particles that has same mass as another particle but opposite charge.
(ii) An anti-matter particle was first discovered during the study of cosmic rays in 1932. Name the anti-particle and give its symbol. positron / anti-electron
(iii)What happens when a particle meets its anti-particle? Pair annihilation occurs and the mass gets converted to energy.
(iv) What is meant by pair production? Pair production involves the production of a particle and its antiparticle from a gamma ray photon.
(v) A photon of frequency 3.6 × 10^20 Hz causes pair production. Calculate the kinetic energy of one of the particles produced, each of which has a rest mass of 9.1 × 10–31^ kg.