Thermal Velocity of Electrons - Higher Physics - Solved Past Paper, Exams of Physics

Download Thermal Velocity of Electrons - Higher Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! PAYS 123) QUESTION 4 [Marks 10] (@) a) Three materials have the energy band siructures shown schematically in the diagram below representing, (1) a metal, (2) an n-type doped semiconductor and (3) an insulator. he shaded ateas indicate occupied (by electrons) energy ranges. @) @) Epo 5eV [t B.=B-lev [a ev Gi) ~~ For the metal shown in (1), find the Fermi velocity and the thermal velocity of the electrons at 300K. (ii) Find the wavelength of EM radiation that will cause a sharp increase in the electrical conductivity of material (2). Gii) By comparing the energy gap values for materials (2) and (3) state, with your reasoning, whether material (3) is expected to be transparent or opaque to visible light at room temperature. (The visible region of the EM spectrum spans the wavelength range A=400nm to A=700nm approx.) (b) Copper (Cu) metal is monovalent (1 conduction electron per atom) and has density p =8.9x10" kgm and atomic mass 63.5. Use this information to estimate the number of occupied electron energy levels in the conduction band of 1 mm of copper at 300K. SOLUTION {a) (i) Fermi velocity v= pete 26.00) ~1.1x10° ms . m, 9.1xL0 kg Thermal velocity van 2k,T m, 2138x107 JK \(300K) ~ 9.110 =1xl0°ms7 6.63x10™}(3x108 i) 3, teal x10°*)(34107) tom E (L:1(Lex107} 8 (iii) Material (2) is opaque because the band gap magnitude permits absorption of all wavelengths in the visible spectrum (400nm-700nm). Material (3) is transparent because the band gap magnitude means visible wavelengths are not ahsorbed. r (b) Copper has number of carrier per m? is n given by n= pS, where p is the density, Ng is Avogadro’s number, M is the atomic mass. ne Na. _{8. Sui wont") ~a.ady10% m since all available states in the conduction band are occupied n is also the number of occupied states (atypical value of t is 0.1-0.5mm giving a stray field less than 1 I) (ii) Whon the conventional current I‘ to is in the +ve x-direction it is found that probe V" is positive (with respect to V*). Is the silicon slice p-type doped or n-type doped? * Electron drift velocity is in the —x direction. * Magnetic force on charge carriers is F,, = q(vxB) in the —y direction * Fy causes —ve charge to accumulate on the V" probe. This accumulation of charge would mean a Hall voltage measured positive from V+ to V- but the measured voltage is the other way around * Therefore, the charge carriers are holes and the doping is p-type. QUESTION 6 [Marks 16] (a) The current-voltage characteristic for an ideal diode in the forward direction is given by I= 1,(ew? -1) (i) Give the meaning of the symbols in this expression. (ii) Provide a labeled sketch graph of the form of the current-voltage characteristic expected for an ideal diode. Gii) Sketch a simple circuit diagram showing a forward-biased diode. Your series circuit should contain the diode, a battery and a single resistor R only, (iv) The diode in the circuit of part (iii) is rated at 100mA maximum current. If the battery voltage is ¢ = 1.5 V, calculate the minimum valuc of R required. (b) The IR communication port on your PDA (c.g. like the Palm Pilot or pocket PC) operates with an infra-red LED. Estimate the band gap of the semiconductor required for the LED in this device. (c) Light of wavelength 4 =620nm (Inm=10°m) and intensity 2.0 Wm™ is shone onto the surface of a photoconductive detector consisting of a rectangular specimen of semiconductor. The detector has an effective area 1,0x100.0 mm”. The mobilities u of the electrons and holes in the semiconductor are 0.001 m?V"'s™ and 1.0m?V"'s7 respectively. Assuming each photon of light striking the semiconductor is absorbed and produces an electron-hole pair, calculate the change in elecirical conductance produced by the illumination. AN + AP AL for AN photoexcited electrons and AP photoexcited holes and AL=V is the volume of the semiconductor detector.] [Hint: the change in conductance is Ag = hor. The conductivity is o= ne where n= SOLUTION (a) (i) In the equation I= (ee - i) the symbols have the meaning Jp is the saturation reverse bias current e (in exponent) is the electron charge AV is the diode bias voliage kp is Boltzmann’s constant T is the temperature in Kelvin (ii) The ideal diode has I-V characteristic : (mA) Vieverse AV (V) (iii) Cireuit for forward biased diode: Marker Note: battery polarity must be correct for forward bias (as shown) to obtain marks (1.5-0.6) Ww) Re - oy) 100x10° (b) [nfra-red LED, take & = 900nm. (Markers Note: N.B. this is for an estimate only. Other IR wavelengths are alsa acceptable.) 6.63x10}(3x108 he .6ax10™y(3x00") =2.2x10" J=14eV Bey 900x107" (c) The photon energy is 34 8 ea © te _(6 Sno" Jon0 ). aoe0" J