Modern Physics Matter Waves

MODERN PHYSICS DUAL NATURE OF LIGHT Experimental phenomena of light reflection, refraction, interference, diffraction are explained only on the basis of wave theory of light. These phenomena verify the wave nature of light. Experimental phenomena of light photoelectric effect and Crompton effect, pair production and positron inhalational can be explained only on the basis of the particle nature of light. These phenomena verify the particle nature of light. It is inferred that light does not have any definite nature, rather its nature depends on its experimental phenomenon. This is known as the dual nature of light. The wave nature and particle nature both can not be possible simultaneously. De-Broglie HYPOTHESIS De Broglie imagined that as light possess both wave and particle nature, similarly matter must also posses both nature, particle as well as wave. De Broglie imagined that despite particle nature of matter, waves must also be associated with material particles. Wave associated with material particles, are defined as matter waves. (i) De Broglie wavelength associated with moving particles If a particle of mass m moving with velocity v 1 p2 momentum of particle p = mv = mv2  2 2m 1 h h h 1 1          with the particles is    v p mv p E 2 mE Kinetic energy of the particle E  2 mE the wave length associated The order of magnitude of wave lengths associated with macroscopic particles is 10 –24 Å. The smallest wavelength whose measurement is possible is that of – rays ( 10–5 Å). This is the reason why the wave nature of macroscopic particles is not observable. The wavelength of matter waves associated with the microscopic particles like electron, proton, neutron, – particle, atom, molecule etc. is of the order of 10–10 m, it is equal to the wavelength of X–rays, which is within the limit of measurement. Hence the wave nature of these particles is observable. (ii) De Broglie wavelength associated with the charged particles Let a charged particle having charge q is accelerated by potential difference V. Kinetic energy of this particle E  1 mv2  qV 2 Momentum of particle p  mv  2mE = 2mqV The De Broglie wavelength associated with charged particle   (iii) h  p h 2mE  h 2 mqV For an Electron me = 9.1 × 10–31 kg, q = 1.6 × 10–19 C, h = 6.62 × 10–34 J–s De Broglie wavelength associated with electron    6.62  10 34 2  9.1  10 3 1  1.6  10 19 V 12 .2 7  10 10 V meter  Potential difference required to stop an electron of wavelength  is V = 1 2.27 o 1 A So   V V 1 50 .6 volt (Å)2 2 (iv) For Proton mp = 1.67 x 10–27 kg De Broglie wavelength associated with proton p= (v) 2 1 .6 7 1 0 2 7 1.6 10 1 9 V ; p = 0 .2 86 10 1 0 V meter  0 .28 6 V Å For Deuteron md = 2 × 1.67 × 10–27 kg, qd = 1.6 × 10–19 C d = (vi) 6 .6 2 1 0 34 6.6 2 1 0 3 4 2  2 1.67 1 0 27 1 .6 1 0 1 9 V = 0 .202 o A V For Particles q = 2 × 1.6 × 10–19 C, m = 4 × 1.67 × 10–27 kg   6.6 2 10 34  19  V = 0 .101 o A V De Broglie Wavelength Associated with Uncharged Particles (i) Kinetic energy of uncharged particle E  1 p2 mv2  2 2m m = mass of particle, v = velocity of particle, p = momentum of particle. (ii) Velocity of uncharged particle v  (iii) Momentum of particle 2E m p  mv  2mE h h h   p mv 2 mE 2 h  eV 2 2m   1.6  10 19 wavelength associated with the particle   Kinetic energy of the particle in terms of its wavelength E  For a neutron mn = 1.67 × 10–27 kg    6 .62  10 3 4 2  1 .6 7  1 0 27 E  h2 2 m 2 0 .28 6  1 0 1 0 E EXPLANATION OF BOHR QUANTIZATION CONDITION According to De Broglie electron revolves round the nucleus in the form of stationary waves (i. e. wave packet) in the similar fashion as stationary waves in a vibrating string. Electron can stay in those circular orbits whose circumference is an integral multiple of De–Broglie wavelength associated with the electron, 2r = n h nh    and 2r = n  mvr  mv 2 This is the Bohr quantizations condition. equivalent straightened orbit meter eV  0.2 86 o A eV E MODERN PHYSICS Ex. Sol. Find the initial momentum of electron if the momentum of electron is changed by pm and the De Broglie wavelength associated with it changes by 0.50 % d d 0 .5 1  1 00  0 .5    and p = pm   1 00 20 0 p  | dp| d h dp h h 1 p pm 1   , differentiating =– 2 =– × =     p = 200 pm  d    p   p 200 Ex. An –particle moves in circular path of radius 0.83 cm in the presence of a magnetic field of 0.25 Wb/m2. Find the De Broglie wavelength associated with the particle. Sol. = Ex. A proton and an –particle are accelerated through same potential difference. Find the ratio of their de- Broglie wavelength. Sol. 6 .62 1 0 3 4 h h = = meter = 0.01 Å p 2 1.6 10 19  0 .2 5  83 10 4 qBr  h  mv h 2mE  h 2 mqV For –particle m =4 m, q = 2e, Ex. Q E  qV   p mp qp mq  mv2   qvB  Q r   For proton mp = m, q = e  1 2 2 A particle of mass m is confined to a narrow tube of length L. (a) Find the wavelengths of the de–Broglie wave which will resonate in the tube. (b) Calculate the corresponding particle momenta, and (c) Calculate the corresponding energies. Sol. (a) The de–Broglie waves will resonate with a node at each end of the tube. Few of the possible resonance forms are as follows :  n  (b) 2L , n=1,2,3..... n Since de–Broglie wavelengths are  n  h pn pn = h  nh , n  1,2, 3.... n 2 L (c) The kinetic energy of the particles are Kn = p 2n n2 h2  2 , 2m 8 L m n = 1, 2, 3,.... 1. Atomic Models : (a) Thomson model : (Plum pudding model) (1) Most of the mass and all the positive charge of an atom is uniformly distributed over the full size of atom (10-10 m). (2) Electrons are studded in this uniform distribution. (3) Failed to explain the large angle scattering -particle scattered by thin foils of matter. (b) Rutherford model : (Nuclear Model) (1) The most of the mass and all the positive charge is concentrated within a size of 10-14 m inside the atom. This concentrated is called the atomic nucleus. (2) The electron revolves around the nucleus under electric interaction between them in circular orbits. (3) An accelerating charge radiates the nucleus spiralling inward and finally fall into the nucleus, which does not happen in an atom. This could not be explained by this model. (c) Bohr atomic model : Bohr adopted Rutherford model of the atom & added some arbitrary conditions. These conditions are known as his postulates (1) The electron in a stable orbit does not radiate energy. (2) A stable orbit is that in which the angular momentum of the electron about nucleus is an integral (n) h h i.e., mvr = n ; n = 1, 2, 3, ..(n  0). 2 2 (3) The electron can absorb or radiate energy only if the electron jumps from a lower to a higher orbit or falls from a higher to lower orbit. (4) The energy emitted or absorbed is a light photon of frequency  and of energy.. E = h  multiple of For hydrogen atom : (z = atomic number = 1) Ln = angular momentum in the nth orbit = n h 2 o rn = radius of nth circular orbit = (0.529 A ) n2   rn  n 2 En = Energy of the electron in the nth orbit = 13.6eV eV 11 13.6  E n  22  nn22 nn Note : Total energy of the electron in an atom is negative, indicating that it is bound. 13.6 eV n2 (5) En2 – En1 = Energy emitted when an electron jumps from n2th orbit to n1th orbit (n2 > n1) Binding Energy (BE)n = E n  1 1 E  13.6 eV   2  2  n n  1 2  E  h ;  = frequency of spectral line emitted. 1 1 1 = wave no, [no. of waves in unit length (1m)] = R  n 2  n 2    1 2  where R = Rydberg's constant, for hydrogen = 1.097 × 107 m-1 MODERN PHYSICS (6) For hydrogen like atom/species of atomic number Z : rnz  2 o Bohr radius 2  Z2 n n   0.529 A  ; E nz   13.6  2 eV Z n   Z Rz = RZ2 ; Rydberg's constant for element of atomic no. Z. Note : If motion of the nucleus is also considered, then  is replaced by . Where  = reduced mass of electron - nucleus system = mM/(m + M) In this case En = (–13.6 eV) 2. Spectral series : (a) Lyman Series : Z2  . n 2 me (Landing orbit n = 1) 1 1  Ultraviolet region v  R  2  2  ; n 2  1 1 n 2  (b) Balmer Series : Visible region (c) Paschan Series : (Landing orbit n = 2) 1 1 v  R  2  2  ; n2  2  2 n2  (Landing orbit n = 3) 1 1  In the mid infrared region v  R  2  2  ; n 2  3  3 n2  (d) Bracket Series : (Landing orbit n = 4) 1 1 In the mid infrared region v  R  2  2  ; n 2  4  4 n2  (e) Pfund Series : (Landing orbit n = 5) 1 1 In far infrared region v  R  2  2  ; n 2  5 5 n2  In all these series n2 = n1 + 1 is the  line = n1 + 2 is the  line = n1 + 3 is the  line .......etc where n1 = Landing orbit 3. Total emission spectral lines From n1 = n to n2 = 1 state = n  n  1 2   n  m  n  m  1   From n1 = n to n2 = m state =  2   4. Excitation Potential of Atom : Excitation potential for quantum jump from n1  n2 = 5. E n 2  E n1 electron charge Ionization energy of hydrogen atom : The energy required to remove an electron from an atom. The energy required to ionize hydrogen atom is = 0 – (– 13.6) = 13.6 eV. 6. Ionization Potential : Potential difference through which a free electron is moved to gain ionization energy = E n electron charge 7. X - RAYS (a) X-rays are produced by bombarding high speed electrons on a target of high atomic weight and high melting point. (b) Short wavelength (0.1 A to 10 A ) electromagnetic radiation. (c) Are produced when a metal anode is bombarded by very high energy electrons (d) Are not affected by electric and matgnetic field. (e) They cause photoelectric emission. o o Characteristic equation eV = h  m e = electron charge; V = accelerating potential  m = maximum frequency of X - radiation (f) Intensity of X - rays depends on number of electrons hitting the target. (g) Cut off wavelength or minimum wavelength, where V (in volts) is the (h) 12400 & A V Continuous spectrum due to retardation of electrons. 8. Characteristic X - rays p.d. applied to the tube  min  For K  ,   9. hc EK  EL For K  ,   hc EL  EM Moseley's law for characteristic spectrum : Frequency of characteristic line v = a(Z – b) Where a, b are constant, for K  line b = 1 Z = atomic number of target  = frequency of characteristic spectrum b = screening constant (for K -series b = 1, L series b = 7.4), a = proportionality constant MODERN PHYSICS Bohr model 10. Moseley's correction 1. For single electron species 1. For many electron species 2. 1 1 E  13.6 Z 2  2  2  eV  n1 n 2  2. 1 2  1 E  13.6  Z  b   2  2  eV  n1 n 2  3. 1 1   RcZ2  2  2   n1 n 2  3. 1 2  1   RcZ2  Z  b   2  2   n1 n 2  4. 1 1 1  RZ 2  2  2   n n  1 2  4. 1 1 2  1  R  Z  b  2  2   n n  1 2  Diffraction of X - ray Diffraction of X- ray take place according to Bragg's law 2d sin = n d = Spacing of crystal plane or lattice constant or distance between adjacent atomic plane  = Bragg's angle or glancing plane  = Diffracting angle n = 1, 2, 3..... 11. For Maximum Wavelength sin  = 1, n = 1 max = 2d So if  > 2d diffraction is not possible i.e., solution of Bragg's equation is not possible. 12. Binding energy = – [Total Mechanical Energy] 13. Velocity of electron in nth orbit for hydrogen atom  14. Series limit means minimum wave length of that series. 15. Binding energy = – [Total Mechanical Energy] 16. Velocity of electron in nth orbit for hydrogen atom  c ; c = speed of light. 137n 17. c ; c = speed of light. 137n Series limit means minimum wave length of that series. 18. Photo Electric Effect The phenomenon of the emission of electrons, when metals are exposed to light (of a certain minimum frequency) is called photo electric effect. 19. Results : (a) Can be explained only on the basis of the quantum theory (concept of photon) (b) Electrons are emitted if the incident light has frequency    0 (threshold frequency) emission of electrons is independent of intensity. The wave length corresponding to  0 is called threshold wave length 0 (c)  0 is different for different metals (d) Number of electrons emitted per second depends on the intensity of the incident light. 20. Einsteins Photo Electric Equation : Photon energy = KEmax of electron + work function h  KE max    = Work function = energy needed by the electron in freeing itself from the atoms of the metal  = h 0 21. Stopping Potential or Cut Off Potential : The minimum value of the retarding potential to prevent electron emission is eVcut off   KE max Note : The number of photons incident on a surface per unit time is called photon flux. 22. Wave Nature of Matter : Beams of electrons and other forms of mattter exhibit wave properties including interference and diffraction with a de Broglie wave length given by   h (wave length of a pratical) p 23. De Broglie wavelength associated with moving particles If a particle of mass m moving with velocity v. Kinetic energy of the particle E  momentum of particle p = mv = 1 p2 mv 2  2 2m 2mE the wave length associated with the praticles is   24. h h   p mv h 2mE De Broglie wavelength associated with the charged particle (a) For an Electron (b) For Proton p   12.27  1010 V 0.286  10 10 V 0.202 (c) For Deuteron d  (d) For  Particle     V m m 0.286 V o A 0.101 V 0.286 o A V o A o A So   1 V