Physics activity


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Physics activity

  1. 1. PHYSICSACTIVITYName: - Swarup Kumar BoroClass: - 12Roll. No.: -16
  2. 2. Activity 1
  3. 3. Davisson and GermerExperiment:A beam of electrons emitted by the ● ●electron gun is made to fall on Nickel V Electron Guncrystal cut along cubical axis at aparticular angle. F CThe scattered beam of electrons isreceived by the detector which can be Arotated at any angle.The energy of the incident beam of θ Фelectrons can be varied by changing the θapplied voltage to the electron gun.Intensity of scattered beam of electrons is Crystal Latticefound to be maximum when angle ofscattering is 50 and the accelerating Nickel Crystalpotential is 54 V. θ + 50 + θ = 180° i.e. θ = 65° Electron diffraction is similar For Ni crystal, lattice spacing to X-ray diffraction. d = 0.91 Å Bragg’s equation 2dsinθ = nλ gives For first principal maximum, n = 1 λ = 1.65 Å
  4. 4. Incident Beam Incident Beam Intensity of scattered beam at 44 V Intensity of scattered beam at 48 V Incident Beam Incident Beam Ф = 50 Intensity of scattered beam at 54 V Intensity of scattered beam at 64 VAccording to de Broglie’s de Broglie wavelength ofhypothesis, h 12.27 Å λ= or λ= moving electron at V = 54 2meV V Volt is 1.67 Å which is in close agreement with 1.65 Å.
  5. 5. Activity 2
  6. 6. FREQUENCY MODULATIONThe process of changing the frequency of a carrier wave inaccordance with the AF signal. Wave Forms Uniform AF signal modulating the Non-uniform AF signal modulating Carrier wave frequency the Carrier wave frequency
  7. 7. Theory: e m= Em cos mt & ec = Ec cos (ct + )The instantaneous frequency ‘’of modulated wave is:  = c + kEm cos mt (where k isproportionality constant which depends on the modulating system)If cos mt = 1, then  = c kEm or  = c  (where  = kEm ismaximum or peak deviation in carrier frequency) Note that  depends on themagnitude of Em and not upon c.Instantaneous value of FM voltage is: e = Ec cos  is given by the following steps:d  =  dt  d  = 2  dt  d  = 2 (c + kEm cos mt ) dtOn integration, we get  = ct + (/ m) sin mt e = Ec cos [ct + (/ m) sin mt] = Ec cos (ct + mf sin mt) where mf = / m is modulation index for FM
  8. 8. Deviation: The amount by which the frequency of the carrier wave is changedfrom its original unmodulated frequency.The rate at which this change occurs is equal to modulating frequency.Modulation Index: M I for F M is the ratio of maximum frequencydeviation to the modulating = / m Observations:1. mf is measured in radians.2. Eqn. for frequency modulated wave is sine of sine function which gives a complex solution whereby the modulated wave consists of a carrier frequency and infinite number of pairs of side bands (Bessel functions).3. In F M, the overall amplitude and hence the total transmitted power remains constant.
  9. 9. Band Width & Bessel Functions
  10. 10. Merits:1. FM is inherently and practically free from noise.2. Noise can be further reduced by increasing .3. FM receivers can further be improved with the help of limiters to remove amplitude changes, if any.4. All the transmitted power is useful in FM.5. Many independent transmitters can be operated on same frequency without interference.Demerits:1. About 10 times wider channel is required by FM as compared to AM.2. Area of reception for FM is much smaller than for AM.3. FM receivers and transmitters are very complex and costly.
  11. 11. Activity 3
  12. 12. gauss’s theorem:The surface integral of the electric field intensity over any closed hypothetical surface(called Gaussian surface) in free space is equal to 1 / ε0 times the net charge enclosed withinthe surface. 1 n ΦE = E . dS = ∑ qi ε 0 i=1 Sproof of gauss’s theorem for spherically Symmetric Surfaces: 1 q E dΦ = E . dS = r . dS n 4πε0 r2 1 q dS dΦ = r .n r dS 4πε0 r2 O• +q r Here, r . n = 1 x 1 cos 0 = 1 1 q dS dΦ = 4πε0 r2 1 q 1 q q ΦE = dΦ = dS = 4π r2 = 4πε0 r2 4πε0 r2 ε0 S S
  13. 13. proof of gauss’s theorem for a closed surface of any Shape: 1 q E dΦ = E . dS = r . dS n 4πε0 r2 r 1 θ n q dS dΦ = r .n 4πε0 r2 Here, r . n = 1 x 1 cos θ dΩ = cos θ +q • q dS cos θ dΦ = 4πε0 r2 q q q ΦE = dΦ = dΩ = 4π = 4πε0 4πε0 ε0 S S
  14. 14. deduction of coulomb’s law from gauss’s Theorem:From Gauss’s law, q ΦE = E . dS = ε0 E SSince E and dS are in the same direction, r dS q O• ΦE = E dS = +q r ε0 S q or ΦE = E dS = ε0 S q q E x 4π r2 = or E= ε0 4πε0 r2 If a charge q0 is placed at a point where E is calculated, then qq0 F= which is Coulomb’s Law. 4πε0 r2
  15. 15. Activity 4
  16. 16. Van de Graff Generator: SS – Large Copper sphereC1, C2 – Combs with sharp points P2P1, P2 – Pulleys to run belt C2HVR – High Voltage Rectifier DM – MotorIS – Insulating StandD – Gas Discharge TubeT - Target T C1 I S HVR P1 M
  17. 17. Principle:Consider two charged conducting spherical shells such that one is smaller and theother is larger. When the smaller one is kept inside the larger one and connectedtogether, charge from the smaller one is transferred to larger shell irrespective of thehigher potential of the larger shell. i.e. The charge resides on the outer surface of theouter shell and the potential of the outer shell increases considerably.Sharp pointed surfaces of a conductor have large surface charge densities and hencethe electric field created by them is very high compared to the dielectric strength ofthe dielectric (air).Therefore air surrounding these conductors get ionized and the like charges arerepelled by the charged pointed conductors causing discharging action known asCorona Discharge or Action of Points. The sprayed charges moving with high speedcause electric wind.Opposite charges are induced on the teeth of collecting comb (conductor) and againopposite charges are induced on the outer surface of the collecting sphere (Dome).
  18. 18. Construction: Van de Graaff Generator consists of a large (about a few metres in radius) copper sphericalshell (S) supported on an insulating stand (IS) which is of several metres high above the ground. A belt made of insulating fabric (silk, rubber, etc.) is made to run over the pulleys (P1, P2 )operated by an electric motor (M) such that it ascends on the side of the combs. Comb (C1) near the lower pulley is connected to High Voltage Rectifier (HVR) whose otherend is earthed. Comb (C2) near the upper pulley is connected to the sphere S through aconducting rod. A tube (T) with the charged particles to be accelerated at its top and the target at thebottom is placed as shown in the figure. The bottom end of the tube is earthed for maintaininglower potential. To avoid the leakage of charges from the sphere, the generator is enclosed in the steel tankfilled with air or nitrogen at very high pressure (15 atmospheres).
  19. 19. Working:Let the positive terminal of the High Voltage Rectifier (HVR) isconnected to the comb (C1). Due to action of points, electric wind iscaused and the positive charges are sprayed on to the belt (silk orrubber). The belt made ascending by electric motor (EM) and pulley(P1) carries these charges in the upward direction.The comb (C2) is induced with the negative charges which are carriedby conduction to inner surface of the collecting sphere (dome) Sthrough a metallic wire which in turn induces positive charges on theouter surface of the dome.The comb (C2) being negatively charged causes electric wind byspraying negative charges due to action of points which neutralize thepositive charges on the belt. Therefore the belt does not carry anycharge back while descending. (Thus the principle of conservation ofcharge is obeyed.)
  20. 20. The process continues for a longer time to store more and morecharges on the sphere and the potential of the sphere increasesconsiderably. When the charge on the sphere is very high, theleakage of charges due to ionization of surrounding air alsoincreases.Maximum potential occurs when the rate of charge carried in bythe belt is equal to the rate at which charge leaks from the shelldue to ionization of air.Now, if the positively charged particles which are to beaccelerated are kept at the top of the tube T, they get accelerateddue to difference in potential (the lower end of the tube isconnected to the earth and hence at the lower potential) and aremade to hit the target for causing nuclear reactions, etc.
  21. 21. Uses:Van de Graaff Generator is used to produce very high potential difference(of the order of several million volts) for accelerating charged particles. The beam of accelerated charged particles are used to trigger nuclear reactions. The beam is used to break atoms for various experiments in Physics. In medicine, such beams are used to treat cancer.It is used for research purposes.
  22. 22. Activity 5
  23. 23. HF Cyclotron: Oscillator S B B D1 D2 + W D1 D2 N WD1, D2 – Dees N, S – Magnetic Pole Pieces W– Window B - Magnetic FieldWorking: Imagining D1 is positive and D2 is negative, the + vely charged particle kept atthe centre and in the gap between the dees get accelerated towards D2. Due toperpendicular magnetic field and according to Fleming’s Left Hand Rule the charge getsdeflected and describes semi-circular path.When it is about to leave D2, D2 becomes + ve and D1 becomes – ve. Therefore theparticle is again accelerated into D1 where it continues to describe the semi-circular path.The process continues till the charge traverses through the whole space in the dees andfinally it comes out with very high speed through the window.
  24. 24. Theory:The magnetic force experienced by the charge provides centripetal force required todescribe circular path. mv2 / r = qvB sin 90 (where m – mass of the charged particle, Bqr q – charge, v – velocity on the path of radius v= – r, B is magnetic field and 90° is the angle m b/n v and B)If t is the time taken by the charge to describe the semi-circular path inside thedee, then πm Time taken inside the dee depends only on the π r t or t = magnetic field and m/q ratio and not on the v Bq = speed of the charge or the radius of the path. If T is the time period of the high frequency oscillator, then for resonance, 2πm T=2t or T = BqIf f is the frequency of the high frequency oscillator (Cyclotron Frequency), then Bq f= 2πm