Class 12 Physics Formula Book PDF

CLASS 12 : PHYSICS FORMULA BOOK ELECTRIC CHARGES AND FIELDS 1 q...

CLASS 12 : PHYSICS FORMULA BOOK ELECTRIC CHARGES AND FIELDS 1 q (ii) At very large distance i.e. r >> a E = 4 πε 0 r 2 k q1q2 1 q1q2 ‰‰ Coulomb’s law : F = = r 2 4 πε r 2 ‰‰ Torque on an electric dipole placed in a uniform    ‰‰ Relative permittivity or dielectric constant : electric field : τ = p × E or τ = pE sin θ ε ‰‰ Potential energy of an electric dipole in a i.e., ε r or K = ε 0 uniform electric field is U = –pE(cosq2 – cosq1) ‰‰ Electric field intensity at a point distant r from where q1 & q1 are initial angle and final angle 1 q a point charge q is E =. between 4 πε 0 r 2   ‰‰ Electric flux φ = E ⋅ dS ‰‰ Electric dipole momentm, ‰‰ Electric field intensity on axial line (end on ‰‰ Gauss’s law : position) of the electric dipole ‰‰ Electric field due to thin infinitely long straight (i) At the point r from the centre of the electric wire of uniform linear charge density l 1 2 pr dipole, E =. λ 4 πε 0 (r 2 − a 2 )2 E= , 2 πε 0r (ii) At very large distance i.e., (r > > a), (i) At a point outside the shell i.e., r > R 2p E= 1 q 4 πε 0r 3 E= 4 πε 0 r 2 ‰‰ Electric field intensity on equatorial line (board (ii) At a point on the shell i.e., r = R on position) of electric dipole 1 q E= (i) At the point at a distance r from the centre 4 πε 0 R 2 1 p (iii) At a point inside the shell i.e., r < R of electric dipole, E =. 4 πε 0 (r 2 + a 2 )3 / 2 E=0 (ii) At very large distance i.e., r > > a, ‰‰ Electric field due to a non conducting solid 1 p sphere of uniform volume charge density r E=. and radius R at a point distant r from the centre 4 πε 0 r 3 of the sphere is given as follows : ‰‰ Electric field intensity at any point due to an (i) At a point outside the sphere i.e., r > R 1 p 1 q electric dipole E = 1 + 3 cos 2 θ E= · 4 πε 0 r 3 4 πε 0 r 2 ‰‰ Electric field intensity due to a charged ring (ii) At a point on the surface of the sphere (i) At a point on its axis at distance r from its i.e., r = R 1 q 1 qr E= · centre, E = 4 πε 0 R 2 4 πε 0 (r + a 2 )3 / 2 2 Physics 1  (iii) At a point inside the sphere i.e., r < R   ‰‰ Relationship between E and V ρr 1 q r   E= = · , for r < R E = −∇V 3ε 0 4 πε 0 R 3 ‰‰ Electric field due to a thin non conducting where infinite sheet of charge with uniformly charge σ ‰‰ Electric potential energy of a system of two surface density s is E = 2ε0 1 q1q2 point charges is U = ‰‰ Electric field between two infinite thin plane 4 πε 0 r12 parallel sheets of uniform surface charge ‰‰ Capacitance of a spherical conductor of radius density s and – s is E = s/e0. R is C = 4pe0R ‰‰ Capacitance of an air filled parallel plate ELECTROSTATIC POTENTIAL AND CAPACITANCE capacitor W ‰‰ Capacitance of an air filled spherical capacitor ‰‰ Electric potential V = q ab ‰‰ Electric potential at a point distant r from a C = 4 πε 0 b−a point charge q is V = q ‰‰ Capacitance of an air filled cylindrical capacitor 4 πε 0r ‰‰ The electric potential at point due to an electric 2 πε 0 L C= dipole  b ln   1 p cos θ  a V= 4 πε 0 r 2 ‰‰ Capacitance of a parallel plate capacitor ‰‰ Electric potential due to a uniformly charged with a dielectric slab of dielectric constant K, spherical shell of uniform surface charge completely filled between the plates of the density s and radius R at a distance r from the capacitor, is given by centre the shell is given as follows : (i) At a point outside the shell i.e., r > R ‰‰ When a dielectric slab of thickness t and 1 q dielectric constant K is introduced between the V= 4 πε 0 r plates, then the capacitance of a parallel plate (ii) At a point on the shell i.e., r = R ε0 A 1 q capacitor is given by C = V=  1 4 πε 0 R d − t  1 −  K (iii) At a point inside the shell i.e., r > R 1 q ‰‰ When a metallic conductor of thickness t is V= introduced between the plates, then capacitance 4 πε 0 R ‰‰ Electric potential due to a non-conducting solid of a parallel plate capacitor is given by sphere of uniform volume charge density r and radius R distant r from the sphere is given as follows : ‰‰ Energy stored in a capacitor : (i) At a point outside the sphere i.e. r > R 2 1 1 1 Q2 1 q U = 2 CV = 2 QV = 2 C V= 4 πε0 r 1 ‰‰ Energy density : u = ε E2 (ii) At a point on the sphere i.e., r = R 2 0 1 q V= 1 1 1 1 4 πε 0 R ‰‰ Capacitors in series : = + +.... + CS C1 C2 Cn (iii) At a point inside the sphere i.e., r < R 1 q( 3R 2 − r 2 ) ‰‰ Capacitors in parallel : CP = C1 + C2 +.... + Cn V= 4 πε 0 2R3 2 Physics  CURRENT ELECTRICITY ‰‰ Relationship between e, V and r ‰‰ Current, I = q or r = R V (ε ) −1 where e emf of a cell, r internal resistance and R t ‰‰ Current density J = I (Electricity, Class 10) is external resistance A P R ‰‰ Wheatstone’s bridge = ‰‰ Drift velocity of electrons is given by Q S   eE ‰‰ Metre bridge or slide metre bridge vd = − τ Sl. m The unknown resistance, R = ‰‰ Relationship between current and drift velocity 100 − l I = nAe vd ‰‰ Comparison of emfs of two cells by using ‰‰ Relationship between current density and drift ε1 l1 potentiometer = velocity ε 2 l2 J = nevd ‰‰ Determination of internal resistance of a cell by l −l  | v | qEτ / m qτ potentiometer r =  1 2 R ‰‰ Mobility, µ = d = E E = m  l2  electric work done ‰‰ Resistance ‰‰ Electric power P = time taken 1 V2 ‰‰ Conductance : G =. P = VI = I 2 R =. R R ‰‰ The resistance of a conductor is (Electricity, Class 10) m l l m R= 2 =ρ where ρ = 2 MOVING CHARGES AND MAGNETISM ne τ A A ne τ ‰‰ Conductivity : ‰‰ Force on a charged particle in a uniform electric   1 ne τ 2  As µ = vd = eτ  field F = qE σ= = = neµ  ρ m E m  ‰‰ Force on a charged particle in a uniform    ‰‰ If the conductor is in the form of wire of length magnetic field F = q ( v × B ) or F = qvB sin θ l and a radius r, then its resistance is ‰‰ Motion of a charged particle in a uniform magnetic field (i) Radius of circular path is ‰‰ If a conductor has mass m, volume V and density d, then its resistance R is (ii) Time period of revolution is 1 qB (Electricity, Class 10) (iii) The frequency is υ = = T 2 πm ‰‰ A cylindrical tube of length l has inner and (iv) The angular frequency is outer radii r1 and r2 respectively. The resistance between its end faces is Bq ‰‰ Cyclotron frequency, υ = ρl 2πm R=. π ( r22 − r12 ) ‰‰ Biot Savart’s law   ‰‰ Relationship between J, s and E  µ 0 Idl sin θ  µ 0 I ( dl × r ) J = sE dB = or dB = 4π r2 4π r3 ‰‰ The resistance of a conductor at temperature ‰‰ The magnetic field B at a point due to a straight t°C is given by Rt = R0 (1 + at + bt2) wire of finite length carrying current I at a ‰‰ Resistors in series Rs = R1 + R2 + R3 perpendicular distance r is 1 1 1 1 ‰‰ Resistors in parallel = + +. µ0I Rp R1 R2 R3 B= [sin α + sin β] 4πr (Electricity, Class 10) Physics 3 ‰‰ The magnetic field at a point on the axis of the ‰‰ If a is the angle between plane of the coil and circular current carrying coil is the magnetic field, then torque on the coil is µ 2πNIa 2 t = NIAB cosa = MB cosa B = 0 2 2 3/2 4π (a +x ) ‰‰ Workdone in rotating the coil through an angle ‰‰ Magnetic field at the centre due to current q from the field direction is carrying circular arc W = MB (1 – cos q) µ0 Iφ ‰‰ Potential energy of a magnetic dipole. B=   4 πa U = − M ⋅ B = − MB cos θ ‰‰ An electron revolving around the central ‰‰ The magnetic field at the centre of a circular nucleus in an atom has a magnetic moment and coil of radius a carrying current I is it is given by µ 0 2 πI µ 0 I B= = 4π a 2a ‰‰ Conversion of galvanometer into a ammeter If the circular coil consists of N turns, then µ 0 2 πNI µ 0 NI  Ig  B= = S= G 4π a 2a  I − Ig    ‰‰ Conversion of galvanometer into voltmeter ∫ ‰‰ Ampere’s circuital law  B ⋅ dl = µ0 I. V ‰‰ Magnetic field due to an infinitely long straight R= −G Ig solid cylindrical wire of radius a, carrying ‰‰ In order to increase the range of voltmeter n current I times the value of resistance to be connected in series with galvanometer is R = (n – 1)G. (a) Magnetic field at a point outside the wire µ0I ‰‰ Magnetic dipole moment   i.e. (r > a) is B = M = m (2l ) 2πr (b) Magnetic field at a point inside the wire ‰‰ The magnetic field due to a bar magnet at any µ Ir i.e. (r < a) is B = 0 2 point on the axial line (end on position) is 2πa µ0 2 Mr (c) Magnetic field at a point on the surface of a B = axial 4π (r 2 − l 2 ) 2 µ0 I For short magnet l2 TC ) t = RC is the time constant of CR circuit. T − TC ‰‰ During discharging of capacitor through resistor ELECTROMAGNETIC INDUCTION q = q0e–t/RC = q0e–t/t ‰‰ Magnetic Flux ALTERNATING CURRENT   φ = B ⋅ A = BA cos θ ‰‰ Mean or average value of alternating current or ‰‰ Faraday’s law of electromagnetic induction dφ voltage over one complete cycle ε=− T dt ‰‰ When a conducting rod of length l is rotated ∫ I0 sin ωt dt I m or I or I av = 0 =0 perpendicular to a uniform magnetic field B, then T induced emf between the ends of the rod is ∫ dt 0 T ∫ V0 sin ωt dt |e| = Bu (pl ) = BuA 2 Vm or V or Vav = 0 T =0 ‰‰ The self induced emf is ∫ dt 0 dφ dI ‰‰ Average value of alternating current for first ε=− =−L dt dt half cycle is ‰‰ Self inductance of a circular coil is T/2 µ N 2 πR ∫ I 0 sin ωt dt L= 0 2I0 2 I av = 0 = = 0.637 I 0 T/2 𠉉 Let IP be the current flowing through primary ∫ dt coil at any instant. If fS is the flux linked with 0 ‰‰ Similarly, for alternating voltage, the average secondary coil then value over first half cycle is fS ∝ IP or fS = MIP Physics 5  T/2 1 ∫ V0 sin ωtdt υr = 2V 2π LC Vav = 0 = 0 = 0.637V0 T/2 π ∫ dt 0 ‰‰ Quality factor ‰‰ Average value of alternating current for second cycle is T ∫ I 0 sin ωtdt T/2 2I0 I av = =− = − 0.637 I 0 T π ∫ dt T/2 ‰‰ Similarly, for alternating voltage, the average ‰‰ Average power (Pav) : value over second half cycle is V0 I 0 T Pav = Vrms I rms cos φ = 2 cos φ ∫ V0 sin ωtdt VI 2V0 ‰‰ Apparent power : Pv = Vrms I rms = 0 0 Vav = T/2 =− = − 0.637 V0 2 T π ∫ dt ‰‰ Efficiency of a transformer, T/2 ‰‰ Mean value or average value of alternating output power VS IS η= =. current over any half cycle input power VP I P 2I0 I av = = 0.637 I 0 π ELECTROMAGNETIC WAVES 2I0 I av = = 0.637 I 0 ‰‰ The displacement current is given by 𠉉 Root mean square (rms) value of alternating current I0 ‰‰ Four Maxwell’s equations are : I rms or I v = = 0.707 I 0 €€ Gauss’s law for electrostatics 2 Similarly, for alternating voltage V0 €€ Gauss’s law for magnetism Vrms = = 0.707 V0 2 ‰‰ €€ Faraday’s law of electromagnetic induction   dφB ‰‰ Inductive reactance : ∫ E ⋅ dl = − dt XL = wL = 2puL €€ Maxwell-Ampere’s circuital law 1 1 ‰‰ Capacitive reactance : XC = = ωC 2 πυC The impedance of the series LCR circuit. ‰‰ The amplitudes of electric and magnetic fields in free space, in electromagnetic waves are ( ) 2 1 Z = R 2 + ( X L − X C ) 2 = R 2 + ωL − related by ωC 1 1 E0 ∴ Admittance = or Y = E0 = cB0 or B0 = Impedance Z c ‰‰ The speed of electromagnetic wave in free 1 space is ∴ Susceptance = Reactance 1 1 c= €€ Inductive susceptance = µ0ε0 Inductive reactance 1 1 ‰‰ The speed of electromagnetic wave in a or SL = = X L ωL medium is 1 €€ Capacitive susceptance = Capacitive reactance 1 1 or SC = = = ωC XC 1/ ωC ‰‰ The energy density of the electric field is ‰‰ The resonant frequency is 1 uE = ε E2 2 0 6 Physics ‰‰ The energy density of magnetic field is ‰‰ Superficial magnification : 1 B2 area of image uB = mS = = m2 2 µ0 area of object ‰‰ Average energy density of the electric field is 1 1 1 ‰‰ Mirror's formula + = 1 u v f < uE > = ε 0 E02 4 ‰‰ Newton’s formula is f 2 = xy, ‰‰ Average energy density of the magnetic field sin i 1 is ‰‰ Laws of refraction : = µ2 2 sin r 1B 1 < uB > = = ε E20 ‰‰ Absolute refractive index : 4 µ0 4 0 0 ‰‰ Average energy density of electromagnetic wave is 1 2 < u > = ε 0 E0 2 ‰‰ Intensity of electromagnetic wave sin (i − r ) Lateral shift, d = t 1 cos r I = < u > c = ε 0 E02 c 2 (Light, Reflection and Refraction, Class 10) ‰‰ Momentum of electromagnetic wave ‰‰ If there is an ink spot at the bottom of a glass U slab, it appears to be raised by a distance p= (complete absorption) c 2U p= (complete reflection ) c ‰‰ The poynting vector is ‰‰ When the object is situated in rarer medium,  1   S= µ0 (E × B) the relation between m1 (refractive index of rarer medium) m2 (refractive index of the RAY OPTICS AND OPTICAL INSTRUMENTS spherical refracting surface) and R (radius of curvature) with the object and image distances ‰‰ When two plane mirrors are inclined at an is given by angle q and an object is placed between them, µ1 µ 2 µ 2 − µ1 the number of images of an object are formed − + = u v R due to multiple reflections. ‰‰ When the object is situated in denser medium, 360° Position of Number of the relation between m1, m2, R, u and v can be n= θ object images obtained by interchanging m1 and m2. In that case, the relation becomes even anywhere n–1 µ 2 µ1 µ1 − µ 2 µ µ µ − µ1 odd symmetric n–1 − + = or − 1 + 2 = 2 u v R v u R asymmetric n ‰‰ Lens maker’s formula 1 1 1  ‰‰ If 360° is a fraction, the number of images = (µ − 1)  −  θ f  R1 R2  formed will be equal to its integral part. ‰‰ Thin lens formula (Light, Class 8) ‰‰ The focal length of a spherical mirror of radius R is given by ‰‰ Linear magnification size of image ( I ) v m= =. size of object (O) u ‰‰ Transverse or linear magnification ‰‰ Power of a lens size of image v m= =− size of object u ‰‰ Longitudinal magnification : ‰‰ Combination of thin lenses in contact dv 1 1 1 1 mL = − = + + +.... du F f1 f 2 f 3 Physics 7 ‰‰ The total power of the combination is given by Length of tube, L = vo + fe P = P1 + P2 + P3 +... ‰‰ When the final image is formed at least distance ‰‰ The total magnification of the combination is of distinct vision, given by m = m1 × m2 × m3.... ‰‰ When two thin lenses of focal lengths f1 and f2 where uo and vo represent the distance of object are placed coaxially and separated by a distance and image from the objective lens, fe is the focal d, the focal length of a combination is given by length of an eye lens. 1 1 1 d = + −.  f D  F f1 f 2 f1 f 2 Length of the tube, L = vo +  e   fe + D  ‰‰ In terms of power P = P1 + P2 – dP1P2. ‰‰ Astronomial telescope (Light, Reflection and Refraction, Class 10) fo magnifying power, M = fe ‰‰ If I1, I2 are the two sizes of image of the object of size O, then O = I1I 2  fD  Length of tube, L = fo +  e ‰‰ The refractive index of the material of the  fe + D  prism is WAVE OPTICS  ( A + δm )  sin  µ=  2  ‰‰ For constructive interference (i.e. formation of sin ( ) A 2 bright fringes) €€ For nth bright fringe, where A is the angle of prism and dm is the angle d of minimum deviation. Path difference = xn = nλ D δV + δ R ‰‰ Mean deviation δ =. where n = 0 for central bright fringe 2 n = 1 for first bright fringe, ‰‰ Dispersive power, n = 2 for second bright fringe and so on angular dispersion (δV − δR ) ω= d = distance between two slits mean deviation (δ) D = distance of slits from the screen µV − µ R ω= , xn = distance of nth bright fringe from the (µ − 1) centre. µ + µR D where µ = V = mean refractive index ∴ xn = nλ 2 d ‰‰ Magnifying power, of simple microscope ‰‰ For destructive interference (i.e. formation of angle subtended by image at the eye dark fringes). M= angle subtended by thee object at the eye €€ For nth dark fringe, tan β β d λ path difference = xn = ( 2n − 1) = = D 2 tan α α where ‰‰ When the image is formed at infinity (far n = 1 for first dark fringe, point), D M= n = 2 for 2nd dark fringe and so on. f xn = distance of nth dark fringe from the centre ‰‰ When the image is formed at the least distance λ D of distinct vision D (near point), ∴ xn = ( 2n −1) 2 d ‰‰ Fringe width, β = λD d ‰‰ Magnifying power of a compound microscope β λ ‰‰ Angular fringe width, θ = = M = mo × me D d ‰‰ If W1, W2 are widths of two slits, I1, I2 are ‰‰ When the final image is formed at infinity intensities of light coming from two slits; a, b are (normal adjustment), the amplitudes of light from these slits, then vo  D  M= W1 I1 a 2 uo  f e  = = W2 I 2 b 2 8 Physics  I max ( a + b)2 (f0), then maximum kinetic energy of the = emitted electron is given as I min ( a − b)2 I − I min Kmax = hu – f0 ‰‰ Fringe visibility V = max For u > u0 or eV0 = hu – f0 = hu – hu0 I max + I min 1 1  ‰‰ When entire apparatus of Young’s double or eV0 = K max = hc  − . slit experiment is immersed in a medium of  λ λ0  refractive index m, then fringe width becomes ‰‰ de Broglie wavelength, λ′D λD β β′ = = = ‰‰ If the rest mass of a particle is m0, its de Broglie d µd µ ‰‰ When a thin transparent plate of thickness t and wavelength is given by 1/ 2 refractive idnex m is placed in the path of one  v2  of the interfering waves, fringe width remains h1 − 2  unaffected but the entire pattern shifts by λ=  c  m0 v D β ‰‰ In terms of kinetic energy K, de Broglie ∆x = (µ −1) t = (µ − 1) t d λ h wavelength is given by λ =. ‰‰ Diffraction due to a single slit 2mK Width of secondary maxima or minima ‰‰ If a particle of charge q is accelerated through a λD λf potential difference V, its de Broglie wavelength β= = where a a h is given by λ =. a = width of slit 2mqV D = distance of screen from the slit V ) ( 150 1/ 2 f = focal length of lens for diffracted light For an electron, λ = Å. ‰‰ For a gas molecule of mass m at temperature ‰‰ Width of central maximum ‰‰ Angular width fringe of central maximum by T kelvin, its de Broglie wavelength is given 2λ h =. λ= , where k is the Boltzmann a 3mkT constant. ‰‰ Angular fringe width of secondary maxima or λ ATOMS minima = a a2 ‰‰ Rutherford’s nuclear model of the atom ‰‰ Fresnel distance, ZF = λ N ntZ 2 e 4 ‰‰ Resolving power of a microscope N (θ) = i (8 πε 0 ) r K 2 sin 4 (θ / 2) 2 2 1 2µ sin θ Resolving power = = The frequency of incident alpha particles d λ ‰‰ Resolving power of a telescope scattered by an angle q or greater 1 D 2 Resolving power = =  Ze 2  2θ dθ 1.22 λ f = πnt  4 πε K  cot 2  0  DUAL NATURE OF RADIATION AND MATTER ‰‰ The scattering angle q of the a particle and hc impact parameter b are related as ‰‰ Energy of a photon E = hυ = λ Ze 2 cot(θ / 2) ‰‰ Momentum of photon is b= E hυ 4 πε0 K p= = c c E hυ ‰‰ Distance of closest approach ‰‰ The moving mass m of photon is m = =. c2 c2 2 Ze 2 ‰‰ Stopping potential r0 = 4 πε0 K 1 K = eV = mv 2 ‰‰ Angular momentum of the electron in a max 0 2 max stationary orbit is an integral multiple of h/2p. ‰‰ Einstein’s photoelectric equation If a light of frequency u is incident on a i.e., L = nh or, mvr = nh photosensitive material having work function 2π 2π Physics 9 ‰‰ The frequency of a radiation from electrons ‰‰ Balmer series makes a transition from higher to lower orbit Emission spectral lines corresponding to the E − E1 transition of electron from higher energy υ= 2 levels (n2 = 3, 4,....∞) to second energy level h ‰‰ Bohr’s formulae (n1 = 2) constitute Balmer series. 1 1 1 (i) Radius of nth orbit = R  2 − 2  λ 2 n  2 where n2 = 3, 4, 5...........,∞ ‰‰ Paschen series (ii) Velocity of electron in the nth orbit Emission spectral lines corresponding to the 1 2 πZe 2 2.2 × 106 Z transition of electron from higher energy vn = = m/s. levels (n2 = 4, 5,.....,∞) to third energy level (n1 4 πε0 nh n = 3) constitute Paschen series. (iii) The kinetic energy of the electron in the nth 1 1 1 1 orbit = − λ R  32 n2   2 ‰‰ Brackett series Emission spectral lines corresponding 13.6 Z2 to the transition of electron from higher = eV. n2 energy levels (n2 = 5, 6, 7,.....,∞) to fourth (iv) The potential energy of electron in nth orbit energy level (n1 = 4) constitute Brackett series. 1 1 1 = R  2 − 2  λ 4 n2  where n2 = 5, 6, 7..........,∞ −27.2 Z2 = eV. ‰‰ Pfund series n2 Emission spectral lines corresponding to the (v) Total energy of electron in nth orbit transition of electron from higher energy levels (n2 = 6, 7, 8,.......,∞) to fifth energy level (n1 = 5) constitute Pfund series. 1 1 1 = R  2 − 2  (vi) Frequency of electron in nth orbit λ 5 n2  2  1  4 π2 Z2 e 4m 6.62 × 1015 Z2 where n2 = 6, 7,...........,∞ υn =   = ‰‰ Number of spectral lines due to transition of  4 πε0  n3h 3 n3 electron from nth orbit to lower orbit is (vii) Wavelength of radiation in the transition n(n − 1) from N=. 2 n2 → n1 is given by 2 13.6 Z ‰‰ Ionization energy = eV. 1  1 1 n2 = RZ2  2 − 2  λ n  1 n2  13.6 Z2 ‰‰ Ionization potential = volt. where R is called Rydberg’s constant. n2 ‰‰ Energy quantisation n2 h 2 En = where n = 1, 2 , 3 ,......... 8mL2 ‰‰ Lyman series NUCLEI Emission spectral lines corresponding to the transition of electron from higher energy ‰‰ Nuclear radius, R = R0A1/3 levels (n2 = 2, 3,...,∞) to first energy level (n1 = where R0 is a constant and A is the mass 1) constitute Lyman series. number 1 1 1 = R  2 − 2  ‰‰ Nuclear density, λ 1 n2  mass nuclear where n2 = 2, 3, 4,......,∞ ρ= volume of nucleus 10 Physics ‰‰ Mass defect is given by ‰‰ The current in the junction diode is given by Dm = [Zmp + (A – Z)mn – mN] I = I0 (eeV/kT –1) ‰‰ The binding energy of nucleus is given by where k = Boltzmann constant, I0 = reverse Eb = Dmc2 = [Zmp + (A – Z)mn – mN]c2 saturation current. = [Zmp + (A – Z)mn – mN] × 931.49 MeV/u. In forward biasing, V is positive and low, ‰‰ The binding energy per nucleon of a nucleus eeV/kT > > 1, then forward current, = Eb/A If = I0 (eeV/kT ) ‰‰ Law of radioactive decay In reverse biasing, V is negative and high dN − λt eeV/kT < < 1, then reverse current, dt = −λN (t) or N (t) = N0 e Ir = – I0 ‰‰ Half-life of a radioactive substance is given by ‰‰ Dynamic resistance ln 2 0.693 ∆V T1 / 2 = λ = λ rd = ∆I ‰‰ Mean life or average life of a radioactive Half wave rectifier substance is given by ‰‰ Peak value of current is 1 T1 / 2 Vm τ= = = 1.44T1 / 2 Im = λ 0.693 rf + RL ‰‰ Activity : R = –dN/dt where rf is the forward diode resistance, RL is ‰‰ Activity law R(t) = R0e–lt the load resistance and Vm is the peak value of where R0 = lN0 is the decay rate at t = 0 and the alternating voltage. R = Nl. ‰‰ rms value of current is ‰‰ Fraction of nuclei left undecayed after n half Im live is Irms = 2 ( ) = ( 21 ) ‰‰ dc value of current is n t /T N 1 1/ 2 or t = nT1/2 = Im N0 2 Idc = ‰‰ Neutron reproduction factor (K) 𠉉 Peak inverse voltage is rate of production of neutrons = P.I.V = Vm rate of loss of neutrons ‰‰ dc value of voltage is Im SEMICONDUCTOR ELETRONICS, MATERIALS, Vdc = Idc RL = R π L DEVICES AND SIMPLE CIRCUITS Full wave rectifier ‰‰ Forbidden energy gap or forbidden band ‰‰ Peak value of current is Vm Eg = hυ = hυ Im = λ rf + RL ‰‰ The intrinsic concentration ni varies with ‰‰ dc value of current is temperature T as 2Im Idc = 3 − Eg / kT π ni2 = A0T e ‰‰ rms value of current is ‰‰ The conductivity of the semiconductor is given Im Irms = by s = e(neme + nhmh) 2 ‰‰ Peak inverse voltage is where me and mh are the electron and hole P.I.V = 2Vm mobilities, ne and nh are the electron and hole ‰‰ dc value of voltage is densities, e is the electronic charge. 2Im ‰‰ The conductivity of an intrinsic semiconductor Vdc = Idc RL = R π L is Ripple frequency si = nie(me + mh) rms value of the components of wave r= ‰‰ The conductivity of n-type semiconductor is average or dc value sn = eNdme 2 ‰‰ The conductivity of p-type semiconductor is I  r =  rms  − 1 sp = eNamh  Idc  Physics 11 ‰‰ For half wave rectifier, Im / 2 π I I Form factor = = = 1.57 Irms = m , Idc = m Im / π 2 2 π 2 ‰‰ For full wave rectifier, I /2 r =  m  −1 Im 2I  Im / π  Irms = , Idc = m 2 π = 1.21 Im / 2 𠉉 For full wave rectifier, Form factor = = = 1.11 2Im / π 2 2 Im 2I Irms = , Idc = m Common emitter amplifier 2 𠉉 dc current gain 2 I / 2  IC r=  m −1 βdc =  2 Im / π  IB ‰‰ ac current gain = 0.482 ∆IC βac = Rectification efficiency ∆IB dc power delivered to load ‰‰ Voltage gain η= ac input power from transformerr secondary Vo R ‰‰ For a half wave rectifier, Av = = − βac × o Vi Ri dc power delivered to the load is ‰‰ Power gain ( )R 2 2 I output power (Po ) Pdc = Idc RL = m L Ap = π input power (Pi ) Input ac power is Vo ‰‰ Voltage gain (in dB) = 20 log10 ( I2 ) (r 2 Vi 2 m Pac = Irms (r f + RL ) = f + RL ) = 20 log10 Av P Rectification efficiency ‰‰ Power gain (in dB) = 10 log o Pi Pdc ( Im / π)2 RL η= = × 100% Common base amplifier Pac ( I / 2)2 (r + R ) m f L ‰‰ dc current gain 40.6 IC = % αdc = 1 + r f / RL IE ‰‰ ac current gain ‰‰ For a full wave rectifier,  ∆I  dc power delivered to the load is αac =  C   ∆IE  ‰‰ Voltage gain ( 2πI ) R 2 2 m Pdc = Idc RL = L Vo R Av = = αac × o Vi Ri Input ac power is ‰‰ Power gain 2 2 I  output power (Po ) Pac = Irms (r f + RL ) =  m  (r f + RL ) Ap =  2 input power (Pi ) Rectification efficiency = aac × Av P ( 2Im / π)2 RL 81.2 ‰‰ Relationship between a and b η = dc = × 100% = % 1 + r f / RL Pac (Im / 2 )2 (r f + RL ) ; If rf