Physics-28..22-Electron & Photon

1. EMISSION OF ELECTRONS At room temperature the free electrons move randomly within the conductor, but they don’t leave the surface of the conductor due to attraction of positive charges. Some external energy is required to emit electrons from a metal surface. Minimum energy is required to emit the electrons which are just on the surface of the conductor. This minimum energy is called the work function( W) . The work function is the property of the metallic surface. The energy required to liberate an electron from metal surface may arise from various source such as heat, light, electric field etc. Depending on the nature of source of energy, the following methods are possible. (i) Thermionic emission: The energy to the free electrons can be given by heating the metal, the electrons so emitted are known as thermions. (ii) Field emission: When a conductor is put under strong electric field the free electrons on it experience an electric force in the opposite direction of field. Beyond a certain limit electrons start coming out of the metal surface. Emission from a metal surface by this method is called the field emission. (iii) Secondary emission: Emission of electrons from a metal surface by the bombardment of high speed electrons or other particle is known as secondary emission. (iv) Photoelectric emission: Emission of free electrons from a metal surface by falling light (or any other electromagnetic wave which has an energy greater than the work function of the metal) is called photoelectric emission. The electrons so emitted are called photoelectrons. This is explained in detail as under. 4.1 PHOTOELECTRIC EFFECT It was observed by Hertz and Lenard around 1880 that when a clean metallic surface is irradiated by monochromatic light of proper frequency, electrons are emitted from it. This phenomenon of ejection of the electrons from metal surface was called as Photoelectric Effect. The electrons thus ejected were called as photoelectrons. For photoemission to take place, energy of incident light photons should be greater than or equal to the work function of the metal. Or E ≥ W hf ≥ W [Where h is plank’s constant] f ≥ Here is the minimum frequency required for the emission of electrons. This is known as threshold frequency fo. Study of photoelectric effect The given set up (as shown in fig.) is used to study the photoelectric effect experimentally. In an evacuated glass tube, two zinc plates C and D are enclosed. Plates C acts as anode and D acts as photosensitive plate. Two plates are connected to a battery B and ammeter A. If the radiation is incident on the plate D through a quartz window W, electrons are ejected out of plate and current flows in the circuit. The plate C can be maintained at desired potential (+ve or -ve) with respect to plate D. With the help of this apparatus, we will now study the dependence of the photoelectric effect on the following factors. 1. Intensity of incident radiation 2. Potenti

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Physics-28..22-Electron & Photon

• 1. 1. EMISSION OF ELECTRONS At room temperature the free electrons move randomly within the conductor, but they don’t leave the surface of the conductor due to attraction of positive charges. Some external energy is required to emit electrons from a metal surface. Minimum energy is required to emit the electrons which are just on the surface of the conductor. This minimum energy is called the work function( W) . The work function is the property of the metallic surface. The energy required to liberate an electron from metal surface may arise from various source such as heat, light, electric field etc. Depending on the nature of source of energy, the following methods are possible. (i) Thermionic emission: The energy to the free electrons can be given by heating the metal, the electrons so emitted are known as thermions. (ii) Field emission: When a conductor is put under strong electric field the free electrons on it experience an electric force in the opposite direction of field. Beyond a certain limit electrons start coming out of the metal surface. Emission from a metal surface by this method is called the field emission. (iii) Secondary emission: Emission of electrons from a metal surface by the bombardment of high speed electrons or other particle is known as secondary emission. (iv) Photoelectric emission: Emission of free electrons from a metal surface by falling light (or any other electromagnetic wave which has an energy greater than the work function of the metal) is called photoelectric emission. The electrons so emitted are called photoelectrons. This is explained in detail as under. 4.1 PHOTOELECTRIC EFFECT It was observed by Hertz and Lenard around 1880 that when a clean metallic surface is irradiated by monochromatic light of proper frequency, electrons are emitted from it. This phenomenon of ejection of the electrons from metal surface was called as Photoelectric Effect. The electrons thus ejected were called as photoelectrons. For photoemission to take place, energy of incident light photons should be greater than or equal to the work function of the metal. Or E ≥ W hf ≥ W [Where h is plank’s constant] f ≥ W h Here W h is the minimum frequency required for the emission of electrons. This is known as threshold frequency fo. Study of photoelectric effect The given set up (as shown in fig.) is used to study the photoelectric effect experimentally. photoelectrons incident radiation A + B e e e e e e C e W D
• 2. In an evacuated glass tube, two zinc plates C and D are enclosed. Plates C acts as anode and D acts as photosensitive plate. Two plates are connected to a battery B and ammeter A. If the radiation is incident on the plate D through a quartz window W, electrons are ejected out of plate and current flows in the circuit. The plate C can be maintained at desired potential (+ve or -ve) with respect to plate D. With the help of this apparatus, we will now study the dependence of the photoelectric effect on the following factors. 1. Intensity of incident radiation 2. Potential difference between C and D 3. Frequency of incident radiation. Effect of Intensity of incident radiation The electrode C i.e. collecting electrode is made positive with respect to D. Keeping the frequency of light and the potentials fixed, the intensity (amount of energy falling per unit area per second) of incident light is varied and the photoelectric current (i) is measured in ammeter. The photoelectric current is directly proportional to the intensity of light. The photoelectric current gives an account of number of photoelectrons ejected per sec. O I A  Effect of p.d. between C & D Keeping the intensity and frequency of light constant, the positive potential of C is increased gradually. The photoelectric current increases with increase in voltage (accelerating voltage) till, for a certain positive potential of plate C, the current becomes maximum beyond which it does not increase for any increase in the accelerating voltage. This maximum value of the current is called as saturation current. I2 >I1 I2 I1 Vc i Low intensity acceleroting potential retarding potential for two diff intensities but freq-same  Make the potential of C as zero and make it increasingly negative. The photoelectric current decrease as the potential is made increasingly negative (retarding potential), till for a sharply defined negative potential Vc of C, the current becomes zero. The retarding potential for which the photoelectric current becomes zero is called as cut- off or stopping potential (Vc). 2 C max 1 eV mv 2 = When light of same frequency is used at higher intensity, the value of saturation current is found to be greater, but the stopping potential remains the same. Hence the stopping potential is independent of intensity of incident light of same frequency. Effect of frequency on Photoelectric Effect  The stopping potential Vc is found to be changing linearly with frequency of incident light being more negative for high frequency. An increase in frequency of the incident light increases the kinetic energy of the emitted electrons, so greater retarding
• 3. potential is required to stop them completely. For a given frequency v, Vc measures the maximum kinetic energy Emax of photoelectrons that can reach plate C. 2 c max 1 eV m V 2 Þ = Where m is the mass of electron, e is charge of electron and Vmax is maximum velocity of electron. This means that work done by stopping potential must just be equal to maximum kinetic energy of an electron.  The effect of changing incident frequency v can also be studied from the plot of photoelectric current Vs potential applied across CD, keeping the intensity of incident radiation same. From graph, we see that imax is same in all cases. (for same intensity). From graph, as v increases, Vc becomes more negative. V'c Vc i + - imax for two diff freq but same intensity Low frequncy potential  The plot of frequency Vs stopping potential for two different metals is shown here. It is clear from graph that there is a minimum frequency f0 and f0' for two metals for which the stopping potential is zero (Vc = 0). So for a frequency below f0 and f0' for two metals, there will not be any emission of photoelectrons. This minimum value of frequency of incident light below which the emission stops, however large the intensity of light may be, is called as threshold frequency. retrading potenital fo: Thershold for metal (1) f'o : Thershold for metal (2) fo f'o f (1) (2) Laws of Photoelectric Emission 1. For a light of any given frequency, photoelectric current is directly proportional to the intensity of light, provided the frequency is above the threshold frequency. 2. For a given material, there is a certain minimum (energy) frequency, called threshold frequency, below which the emission of photoelectrons stops completely, no matter how high is the intensity of incident light. 3. The maximum kinetic energy of the photoelectrons is found to increase with increase in the frequency of incident light, provided the frequency exceeds the threshold limit. The maximum kinetic energy is independent of the intensity of light.
• 4. 4. The photo-emission is an instantaneous process. After the radiation strikes the metal surface, it just takes 10–9 s for the ejection of photoelectrons. 4.2 SOURCE OF RADIATION If P (in Watts) be the power of the source of radiation and  be the intensity of light radiation at a distance R from the source (R is the perpendicular distance of the receiving surface from the source), then: 2 P 4 R I = p If radiation is falling on a plate of area A, then, energy absorbed/sec by the plate is given as: 2 P Energy / sec A A J/ sec 4 R æ ö = I ´ = ´ ç ÷ è p ø If v be the frequency of the source (i.e. light radiation), then energy per photon = hv energy per second number of photons falling on plate per second energy per photon Þ = 2 P × A hc 4 R n/sec = Energy/photon = hv = hv          If 0 v v ³ (v0: threshold frequency of plate) and photon efficiency of plate be x %, then the number of photoelectron emitted per second is: 2 P × A x 4 R n/sec = × hv 100               4.3 HOW TO DETERMINE THE PHOTOELECTRIC CURRENT? Let P be the power of a point source of electromagnetic radiations, then intensity  at distance r from the source is given by ( ) 2 2 P W /m 4 r I = p If A is the area of a metal surface on which radiations are incident, then the power received by the plate is ( ) 2 P P' =ΙA = A W 4 r p If f is the frequency of radiation, then the energy of photon is given by E = hf Area (A) r The number of photons incident on the plate per second (called photon flux) is given by
• 5. 2 P × A P' 4 r = = E hf é ù ê ú p F ê ú ê ú ë û If 0 f f > (threshold frequency) and photon efficiency of the metal plate is %, then the number of photoelectrons emitted per second is given by 2 P A 4 r n 100 hf 100 é ù ´ ê ú F h h p = ê ú ê ú ë û Finally, the photocurrent i is given by i = ne Where e is the charge of an electron ( ) 19 e 1.6 10 J - = ´ Illustration 1: Photoelectric threshold of metallic silver is  = 3800 Å. Ultra-violet light of  = 2600 Å is incident on silver surface. Calculate (i) the value of work function in joule and in eV. (ii) maximum kinetic energy of the emitted photoelectrons. (iii) the maximum velocity of the photo electrons. (mass of the electron = 9.11 x 10–31 kg) Solution: (i) l = 0 3800 Å 34 8 0 10 0 c 6.633 10 3 10 W hf h 3800 10 - - ´ ´ ´ = = = l ´ 34 8 19 10 6.63 10 3 10 J J 5.23 10 J 3.27 eV 3800 10 - - - ´ ´ ´ = = ´ = ´ (ii) Incident wavelength  = 2600 Å - ´ = = ´ 8 10 3 10 f incident frequency Hz 2600 10 Then = - max 0 T hf W - - - ´ ´ ´ ´ = = ´ = ´ = ´ 34 8 19 19 10 6.63 10 3 10 6.63 3 hf 10 J 7.65 10 J 4.78 eV 2600 10 2.6 = - = - = max 0 T hf W 4.78 eV 3.27 eV 1.51eV. (iii) = 2 max max 1 T mv 2 - - - ´ ´ = = = ´ ´ 19 6 1 max max 31 2T 2 2.242 10 v 0.7289 10 ms m 9.11 10 Illustration 2: Sun gives light at the rate of 1400 Wm–2 of area perpendicular to the direction of light. Assume  (sun light) = 6000 Å. Calculate the (a) number of photons/sec arriving at 1m2 area at that part of the earth, and (b) number of photons emitted from the sun/sec assuming the average radius of Earth's orbit is 1.49 x 1011 m. Solution: 2 1400 W /m ; 6000 Å I = l =
• 6. (a) ( ) ( ) 10 21 34 8 1400 1 6000 10 A n/ sec 4.22 10 E/photon 6.63 10 3 10 - - ´ ´ ´ I = = = ´ ´ ´ ´ (b) ( ) ( ) ( ) 2 10 45 34 8 4 R 6000 10 Power of Sun W n/ sec 1.178 10 E/photon 6.63 10 3 10 - - I ´ p ´ ´ = = = ´ ´ ´ ´ Illustration 3: An isolated hydrogen atom emits a photon of 10.2 eV. Calculate: (a) momentum of photon emitted (b) recoil momentum of the atom (c) kinetic energy of the recoil atom. Mass of proton = Mp = 1.67 x 10–27 kg. Solution: (a) momentum (p) = E/c 19 27 8 10.2 1.6 10 p 5.44 10 kg m/ s 3 10 - - ´ ´ Þ = = ´ ´ (b) recoil momentum of atom = p (Law of conservation of momentum) Þ recoil momentum = 5.44 x 10–27 kg m/s (c) KE = 1/2 mv2 (v = recoil speed of atom, m = mass of H-atom) 2 2 1 p p KE m 2 m 2m æ ö Þ = = ç ÷ è ø ( ) ( ) 2 27 27 27 KE 5.44 10 / 2 1.67 10 8.86 10 J - - - Þ = ´ ´ ´ = ´ 2. BOHR MODEL The main features of Rutherford's model, viz. the nucleus and the electrons orbiting it under the action of the Coulomb's law of electrostatic attraction, were retained in Bohr's theory. In addition Bohr introduced the concept of 'radiation less orbits' or 'stationary states' in which the electron revolves around the nucleus, but does not radiate contrary to the laws of electromagnetism. This was a hypothesis, but at least a working one. Radiation occurred only when an electron made a transition from one stationary state to another. The difference between the energies of the two states was radiated as a single photon. Absorption occurred when a transition occurred from a lower stationary state to a higher stationary state. The third principle invoked by Bohr was the correspondence principle: i.e. the spectrum is continuous and the frequency of light emitted equals the frequency of the electron. In the classical limit. Using these general arguments and the existing body of knowledge, Bohr postulated the following: (i) The electron in an atom has only certain definite stationary states of motion allowed to it, called as energy levels. Each energy level has a definite energy associated with it. In each of these energy levels, electrons move in circular orbit around the positive nucleus. The necessary centripetal force is provided by the electrostatic attraction of the protons in the nucleus. As one moves away from the nucleus, the energy of the states increases. (ii) These states of allowed electronic motion are those in which the angular momentum of an electron is an integral multiple of h/2 or one can say that the angular momentum of an electron is quantised. Þ Angular momentum = m v r = n (h/2) Where m is the mass, v is the velocity, r is the radius of the orbit, h is Planck's constant and n is a positive integer.
• 7. (iii) When an atom is in one of these states, it does not radiate any energy but whenever there is a transition from one state to other, energy is emitted or absorbed depending upon the nature of transition. (iv) When an electron jumps from higher energy state to the lower energy state, it emits radiations in form of photons or quanta. However, when an electron moves from lower energy state to a higher state, energy is absorbed, again in form of photons. (v) The energy of a photon emitted or absorbed is given by using Planck's relation (E = hv). If E1 be the energy of any lower energy state and E2 be the energy of any higher energy state then the energy of the photon (emitted or absorbed) is given as E: E = E2 – E1 = h v Where h = Planck's constant and v = frequency of radiation emitted or absorbed. Note: Bohr's Model is applicable only to one-electron atoms like: He+ , Li++ apart from H-atom. In the following section we will define the radius of orbits around the nucleus (Bohr's orbits), the velocity and energy of an electron in various orbits around the nucleus. We will take rn : radius of nth orbit vn : velocity of electron in nth state (orbit) En : energy of nth state m : mass of an electron (9.1 x 10–31 Kg) Z : atomic number (No. of Protons) K = 1/(40) = constant = 9 x 109 N m2 C–2 h : Planck's constant (6.67 x 10–34 J-s) c : velocity of light (3 x 108 m/s) R : Rydberg constant (1.097 x 107 m–1 ) e : Charge on an electron (1.6 x 10–19 C) v : frequency of the radiation emitted or absorbed v : wave number of the spectral line in the atomic spectra From first postulate 2 2 2 n n mv K Ze r r = From second postulate n n nh mv r 2 = p Solving for n r and n v , we have: 2 2 n 2 2 n h radius r 4 Kme Z = = p 2 10 n 0.53 10 m Z - = ´ 2 n 0.53 A Z = 2 n 2 K Ze velocity v nh p = =
• 8. 6 Z 2.165 10 m/s n = ´  The energy of an electron in nth state En is given by: En = KE + PE 2 2 2 n n n 1 K Ze 1 K Ze E mv 2 r 2 r æ ö = + - = - ç ÷ è ø ( ) 2 2 4 2 n n 2 2 2 K e mZ E putting vaue of r n h - p Þ = 2 18 2 Z 2.178 10 J/atom n - = - ´ 2 2 Z 13.6 eV /atom n = - 2 2 Z 1312 kJ/mol n = -  When an electron jumps from an outer orbit (higher energy) n2 to an inner orbit (lower energy) n1, then the energy emitted in form of radiation is given by: 2 1 2 4 2 2 n n 2 2 2 1 2 2 e mZ K 1 1 E E E h n n æ ö p D = = = - ç ÷ è ø As we know that E = hv, c = v and v 1/ = l E v hc D Þ = 2 4 2 2 3 2 2 1 2 1 2 e mZ K 1 1 v ch n n æ ö p Þ = = - ç ÷ l è ø Now this can be represented as: 2 2 2 1 2 1 1 v RZ n n æ ö = - ç ÷ è ø Where 2 4 2 3 2 e K m R ch p = Note: This relation exactly matches with the empirical relation given by Balmer & Rydberg to account for the spectral lines in H-atom spectra. In fact the value of Rydberg constant in the empirical relation is approximately the same as calculated from the above relation (Bohr's Theory). This was the main success of Bohr's Theory i.e. to account for the experimental observations by postulating a theory. 5.1 ENERGY LEVELS OF HYDROGEN ATOM The spectrum of H-atom studied by Lyman, Balmer, Paschen, Bracken and Pfund can now be explained on the basis of Bohr's Model. It is now clear that when an electron jumps from a higher energy state to a lower energy state, the radiation is emitted in form of photons. The radiation emitted in such a transition corresponds to the spectral line in the atomic spectra of H-atom.
• 9. Lyman series Balmer series Pasher Series Braken series Prilind Series Ground State Emission lines higher lower hv E E   n = 6 n = 5 n = 4 n = 3 n = 2 n = 1 1 13.6 E ev 9   2 13.6 E ev 4   Absorptions lines and exciatation energies n   ionization energy Lyman Series When an electron jumps from any of the higher states to the ground state or 1st state (n = 1), the series of spectral lines emitted lies in ultra-violet region and are called as Lyman Series. The wavelength (or wave number) of any line of the series can be given by using the relation: 2 2 2 2 2 1 1 v RZ n 2, 3, 4, 5,..... 1 n æ ö = - = ç ÷ è ø (For H atom Z = 1) Series limit (for H - atom): 1i.e. v R ¥ ® =  line: 2  1; also known as first line or first member  line: 3  2; also known as second line or second member  line: 4  1; also known as third line or third member Balmer Series When an electron jumps from any of the higher states to the state with n = 2 (IInd state), the series of spectral lines emitted lies in visible region and are called as Balmer Series. The wave number of any spectral line can be given by using the relation: 2 2 2 2 2 1 1 v RZ n 3, 4, 5, 6,........... 2 n æ ö = - = ç ÷ è ø Series limit (for H – atom) : R 2 i.e. v 4 ¥ ® = line : 3 2; line : 4 2 ; line : 5 2 a ® b ® g ® Paschen Series When an electron jumps from any of the higher states to the state with n = 3 (IIIrd state), the series of spectral lines emitted lies in near infra-red region and are called as Paschen Series. The wave number of any spectral line can be given by using the relation: 2 2 2 2 2 1 1 v RZ n 4, 5, 6, 7,.............. 3 n æ ö = - = ç ÷ è ø Series limit (for H – atom) : R 3 i.e. v 9 ¥ ® = line : 4 3 ; line : 5 3 ; line : 6 3 a ® b ® g ® Brackett Series
• 10. When an electron jumps from any of the higher states to the state with n = 4 (IVth state), the series of spectral lines emitted lies in far infra-red region and called as Brackett Series. The wave number of any spectral line can be given by using the relation: 2 2 2 2 2 1 1 v R Z n 5, 6, 7, 8,........... 4 n æ ö = - = ç ÷ è ø Pfund Series When an electron jumps from any of the higher states to the state with n = 5 (Vth state), the series of spectral lines emitted lies in far infra-red region and are called as Pfund Series. The wave number of any spectral line can be given by using the relation: 2 2 2 2 2 1 1 v R Z n 6, 7, 8,........... 5 n æ ö = - = ç ÷ è ø Short Review of formulas (for one electron atom or ions): 1. Velocity of electron in nth orbit = vn = 2.165 x 106 Z/n m/s 2. Radius of nth orbit = rn = 0.53 x 10–10 n2 /Z m 3. Binding energy of an electron in nth state = En = –13.6 Z2 /n2 eV/atom 2 2 18 n 2 2 Z Z E 2.17 10 J/atom 13.6 eV /atom n n - = - ´ = - 4. Kinetic energy 2 2 n n KE 1/ 2 mv K Ze /2r = = = 5. Potential energy 2 n PE k Ze /r = = - 6. Total energy of an electron 2 n n E k Ze /2r = - = - PE = 2TE ; PE = –2KE ; TE = –KE 7. Binding energy of an electron in nth state 2 n 2 13.6 E Z eV n = - 8. Ionisation Energy = – B.E. 2 2 13.6 .E. Z eV n I = + 9. Ionisation Potential Ionisation potential 2 2 .E. 13.6 Z V e n I = = 10. Excitation Energy The energy taken up by an electron to move from lower energy level to higher energy level. Generally it defined from ground state.  Ist excitation energy = transition from n1 = 1 to n2 = 2  IInd excitation energy = transition from n1 = 1 to n2 = 3  IIIrd excitation energy = transition from n1 = 1 to n2 = 4 and so on ....  The energy level n = 2 is also called as Ist excited state.  The energy level n = 3 is also called as IInd excited state. & so on ... In general, excitation energy (E) when an electron is excited from a lower state n1 to any higher state n2 is given as:
• 11. 2 2 2 1 2 1 1 E 13.6 Z eV n n æ ö D = - ç ÷ è ø 11. Energy released when an electron jumps from a higher energy level (n2) to a lower energy level (n1) is given as: 2 2 2 1 2 1 E 13.6 Z eV n n æ ö D = ç ÷ - è ø If v be the frequency of photon emitted and  be the wavelength, then: c E hv h D = = l The wavelength () of the light emitted an also be determined by using: 2 2 2 1 2 1 1 1 v R Z n n æ ö = = - ç ÷ l è ø R = 1.096 x 107 /m Important: Also remember the value of 1/R = 911.5 Å for calculation of  to be used in objectives only). 12. The number of spectral lines when an electron falls from n2 to n1 = 1 (i.e. to the ground state) is given by: ( ) 2 2 n n 1 No. of lines 2 - = If the electron falls from n2 to n1, then the number of spectral lines is given by: ( )( ) 2 1 2 1 n n 1 n n No. of lines 2 - + - = Illustration 4: A doubly ionised Lithium atom is hydrogen like with atomic number 3. (i) Find the wavelength of radiation required to excite the electron in Li++ from the first to the third Bohr Orbit. (Ionization energy of the hydrogen atom equals 13.6 eV). (ii) How many spectral lines are observed in the emission spectrum of the above excited system? Solution: (i) - = 2 n 2 13.6 Z E n ( ) ( )[ ] ( )( ) é ù = D = - = - ´ - ê ú ë û = + ´ - = ´ = 2 3 1 2 2 1 1 Excitation energy E E E 13.6 3 3 1 13.6 9 1 1/ 9 13.6 9 8 / 9 108.8 eV. ( )( ) ( )( ) ( ) ( )( )( ) - - - - ´ ´ l = = D ´ ´ æ ö ´ = ç ÷ è ø = ´ = 34 8 19 7 10 6.63 10 3 10 hc Wavelength E 13.6 8 1.6 10 6.63 3 10 13.6 8 1.6 114.26 10 m 114.3 Å
• 12. (ii) From the excited state (E3), coming back to ground state, there can be 3 C2 = 3 possible radiations. Illustration 5: Radiation emitted in the transition n = 3 to n = 2 orbit in a hydrogen atom falls on a metal to produce photoelectrons, the electrons emitted from the metal surface with the maximum kinetic energy are allowed to pass through a magnetic field of strength 3.125x10–4 T and it is observed that that the paths haves radius of curvature 10 mm Find (a) the kinetic energy of the fastest photo electrons. (b) the work function of the metal. (c) the wavelength of the radiation. Solution: The kinetic energy of the fastest photoelectrons is given by 23 2 2 1 1 E hv 13.6 eV 2 3           The Kinetic energy of the fastest photoelectrons is given by 2 mv Bev r  or mv = p = Ber  Kinetic Energy = 2 2 2 2 1 B e r mv 2 2m  =     2 4 19 4 31 3.125 10 1.6 10 10 J 2 9 10           = 0.86 eV (a) The kinetic energy of the fastest electron is 0.86 eV (b) The work function,  is given by  = (1.9 – 0.86) eV = 1.04 eV (c) The wavelength of the emitted radiation is,     34 8 19 6.63 10 3 10 hc m E 1.3 1.6 10            = 6.54 × 10–7 m = 6540Å 3. DE-BROGLIE WAVES In 1925, before the discovery of electron diffraction, de Broglie proposed that the wavelength () of waves associated with particles (like Electron, photons) of momentum 'p' is given by h h p mv    The wavelength associated with an electron accelerated through a potential difference of V volt is given by = = 2 e e 1 2 eV m v eV or v 2 m l = = e e h h m v 2 eVm Illustration 6: A particle of mass m is confined to a narrow tube of length L. Find
• 13. (a) the wavelengths of the de-Broglie wave which will resonate in the tube, (b) the corresponding particle momenta, and (c) the corresponding energies. Solution: L N A N A A N N A A A N N N N L ( / 2)   L 2( /2)   L 3( / 2)   (a) The de Broglie waves will resonate with a node at each end of the tube. A few of the possible resonance forms are as follows: l = = n 2L ; n 1, 2, 3,...... n (b) Since de-Broglie wavelengths are l = n n h p = = = l n n h nh p n 1, 2, 3,...... 2L (c) ( ) = = = 2 2 2 n 2 n p n h KE , n 1, 2, 3,...... 2m 8 L m Illustration 7: What is the energy and wavelength of a thermal neutron? Solution: By definition, a thermal neutron is a free neutron in a neutron gas at about 20C (293 K). ( )( ) - - = = ´ = ´ 23 21 3 3 KE kT 1.38 10 293 6.07 10 J 2 2 ( ) ( )( ) - - - ´ l = = = = ´ ´ 34 27 21 0 h h 6.63 10 0.147 nm p 2m KE 2 1.67 10 6.07 10 4. X-RAYS X-rays were discovered accidentally by Rontgen in 1895. The first Nobel Prize was awarded to Rontgen in 1901. This highly penetrating electromagnetic radiation has proved to be a very powerful tool to study the crystal structure, in material research, in the radiography of metals and in medical sciences. Laue, Henry and Lawrence Bragg, Barkla, Siegbahn were some of the Nobel Laureates who have made contribution to these studied.  ray spectroscopy and electron-spectroscopy were some of the spin-offs of these studies apart from the discovery of elements. Experimental production of X-rays and the Bragg spectrometer: Electrons from a a heated element were accelerated by very high potential and made to impinge on the target (anode). The X-rays produced are collimated by parallel plates and are incident on a crystal (LiF, quartz, diamond, etc.) As the inter-atomic distance is of the same order as the wavelength of X-rays diffraction is produced and they are detected by counters or photographic plates.
• 14. X – Rays has following property (i) Short wavelength (0.1 A to 1 A) electromagnetic radiation. (ii) Are produced when a metal anode is bombarded by very high energy electrons. (iii) Are not affected by electric and magnetic field. (iv) They cause photoelectric emission. Characteristics equation eV = hvm e = electron charge; V = accelerating potential vm = maximum frequency of X radiation (v) Intensity of X – rays depends on number of electrons hitting the target. (vi) Cut off wavelength or minimum wavelength, where v(in volts) is the p.d. applied to the tube min 12400 V l @ A. I Continous Specutrum 35000 volt K ,K   Characternsic Specutrum  w K K (vii) Continuous spectrum due to retardation of electrons. (viii) Characteristic Spectrum due to transition of electron from higher to lower ( ) ( ) [ ] 2 2 v z b ; a z b Moseley's Law µ - u = - b = 1 for K ; B = 7.4 for L Where b is Shielding factor (different for different series). (ix) Bragg's Law 2 d sin  = n  ( = angle for max intensity) Note: (a) Binding energy = – [Total Mechanical Energy] (b) Vel. of electron in nth orbit for hydrogen atom c 137n @ ; c = speed of light. (c) For x – rays ( ) 2 2 2 1 2 1 1 1 R z b n n æ ö = - - ç ÷ l è ø (d) Series limit of series means minimum wave length of that series. 7.1 THEORETICAL EXPLANATION Production of Continuous Spectrum The accelerated electrons are suddenly stopped by the target. According to Maxwell's theory, wherever a charged particle is accelerated or decelerated, they emit radiation. This is called Bremsstrahlung or braking radiation.
• 15. When the whole of the energy of the electron is converted to radiation, one gets the maximum energy or minimum. If V is the potential difference applied, it is converted to the kinetic energy of the electron. If Ve = E = hv, then the wave-length hc E   . hc = 12400 eVÅ, and if E is given in eV, min. hc eV.A . E(eV)          min.  is directly given in Å. When the electrons lose their energy by multiple collisions and penetration inside the target, the radiation, produced has less energy The energy that is not converted to radiation only heats up the target. It has to be cooled property to prevent damage to the X-ray tube. The maximum value of intensity is approximately at min. 3 2  (This is often marked min.  Actually it is not the maximum value of  but  at intensity maximum). The distribution of intensity depends on the material of the target, the current and the potential difference applied. I  V2 and I  current The current is normally in milli-amperes and the voltage in kilovolts. Illustration 8: If the wavelength of the K line of platinum is 0.2Å, what is the energy needed to excite this line? (the energy of the first level of platinum, n 1 E  = 81keV). What is the corresponding absorption energy? What is the ionization energy of platinum? Solution: The energy corresponding to Q.2Å is 12400eV A E 0.2A      = 62 × 103 eV = 62 KeV But if one gives 62 keV energy to the electron in the x-ray tube, one cannot et K x-ray line because the higher levels are full. K line can be excited only when a vacancy is created in the ground state, i.e to remove the electron from the k-level. The energy required is obviously the ionization energy of the k-level. which is 81 keV. There is no absorption line corresponding to the K emission line. In x-rays one gets only absorption edge (see the theory part). The ionization energy of the k-level is same as the k-absorption edge which is the energy of excitation of the k-electron. This is 81 keV for platinum. Illustration 9: An X-ray tube, operated at a potential difference of 40 kV, produces heat at the rate of 720 W. Assuming 0.5% of the energy of the incident electrons is converted into X-rays. Calculate (i) the number of electrons per second striking the target. (ii) the velocity of the incident electrons. Solution: (i) Heat produced per second at the target is P = 0.995 VI ( 0.5% of energy is converted into X-rays) ( )( ) I = = = ´ 3 P 720 0.018 A 0.995 V 0.995 40 10 The number of electrons per second incident on the target
• 16. - I = = = ´ ´ 17 19 0.018 n 1.1 10 electrons e 1.6 10 (ii) Energy of incident electrons = 2 1 mv eV 2 = = ´ 8 2 eV or v 1.2 10 m/s m 5. ASSIGNMENT 1. The electron in a hydrogen atom make a transition from an excited state to the ground state. Which of the following statements is true? (A) Its kinetic energy increases and its potential and total energy decrease. (B) Its kinetic energy decreases, potential energy increases and its total energy remains the same. (C) Its kinetic and total energies decreases and its potential energy increases. (D) its kinetic, potential and total energies decreases. Solution: (A) The K.E and P.E of an orbital electron in a hydrogen atom is given by (K.E.) = 4 2 2 2 0 me 8 h n  (P.E.) = 4 2 2 2 0 me 8 h n  During the transition from an excited state to lower orbits, n decreases and the K.E. increases and the potential energy become, more negative i.e. it decrease? Similarly the total energy becomes more negative. 2. A photoelectric cell is illuminated by a small bright source of light placed at 1m. If the same source of light is placed 2m away, the electrons emitted by the cathode (A) each carries one quarter of its previous momentum. (B) each carries one quarter of its previous energy. (C) are half the previous number. (D) are one quarter of the previous number. Solution: (D) 2 1 d I µ On doubling the distance the intensity becomes one fourth i.e. only one fourth of photons now strike the target in comparison to the previous number. Since photoelectric effect is a one photon-one electron phenomena, so only one-fourth photoelectrons are emitted out of the target hence reducing the current to one fourth the previous value. 3. As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doublyionzed Li atom (Z = 3) is: (A) 1.51 (B) 13.6 (C) 40.8 (D) 122.4. Solution: (D) According to Bohr's theory, 2 2 n 2 2 RhcZ 13.6z E eV n n    Here z = 3, n = 1  Required energy = |En| = 13.6 × 9
• 17. = 122.4 4. A hydrogen atom and a Li++ ion are both in the second excited state. If H and Li are their respective electronic angular momenta, and EH and ELi their respective energies, then: (A)    H Li and |EH| >|ELi| (B)    H Li and |EH| < |ELi| (C)    H Li and |EH| >|ELi| (D)    H Li and |EH| < |ELi| Solution: (B) The angular momentum  = h n 2 And the total energy E = 2 4 2 2 0 mz e 8 h   Thus, H Li,    but H Li |E | | E | = 2 1 1 Z   |EH| < |ELi| 5. A proton, a deutron and an alpha particle are accelerated through potentials of V, 2V and 4V respectively. Their velocities will bear a ratio. (A) 1 : 1 : 1 (B) 1 : 2 : 1 (C) 2 : 1 : 1 (D) 1 : 1 : 2 Solution: (D) 2qV v m = 6. The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately (A) 540 nm (B) 400 nm (C) 310 nm (D) 220nm Solution: (C) Work Function 0  = hv0 = hv0 = max. hc   max. 0 hc 1240eV.nm 310nm 4eV      7. An electron is 2000 times lighter than a proton. Both are moving such that their matter waves have a length of 1Å. The ratio of their kinetic energy in approximation is (A) 1 : 1 (B) 1 : 2000 (C) 2000 : 1 (D) 1 : 200 Solution: (C) Since both have same de Broglie wavelength, hence both must have equal value of momentum. Since 2 p E 2m = p e p e m E 1840 2000 E m Þ = = nearest possible approximation to answer ì ü ï ï í ý ï ï î þ 8. An electron is lying initially in the n = 4 excited state. The electron de-excites itself to go to n = 1 state directly emitting a photon of frequency v41. If the same electron first
• 18. de-excites to n = 3 state by emitting a photon of frequency v43 and then goes from n = 3 to n = 1 state by emitting a photon of frequency v31, then (A) v41 = v43 + v31 (B) v41 = v43 – v31 (C) v43 = v41 + 2v31 (D) Data Insufficient Solution: (A) According to Ritz Combination Principle m n m i i n v v v ® ® ® = + (for m < i < n) e.g. 4 1 4 3 3 1 v v v or ® ® ® = + 4 1 4 2 2 1 v v v ® ® ® = + 9. The K X-ray emission line of tungsten occurs at  = 0.021 nm. The energy difference between K and L levels in this atom is about (A) 0.51 MeV (B) 1.2 MeV (C) 59 MeV (D) 13.6 MeV Solution: (C) hc E D = l 10. The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6eV fall on it is 2eV. The stopping potential in volts is: (A) 2 (B) 4 (C) 6 (D) 10 Solution: (A) According to Einstein's photoelectric equation hv = 0  + (KE)max = 0  +eVs, where eVs = (K.E.) max.  Slopping potential Vs =  max. KE 2eV 2V e e   11. An electron enters electric field of 104 B/m perpendicular to the field with a velocity of 107 ms–1 . The vertical displacement of electron after one mille second will be : (A) 8 8 10 m 91 ´ (B) 8 91 10 m 8 ´ (C) 10 8 10 m 91 ´ (D) 10 91 10 m 8 ´ Sol. The vertical displacement of electron 2 2 1 1 qE y at t 2 2 m = = or ( ) 2 19 4 3 10 31 1.6 10 10 10 8 y 10 m 2 9.1 10 91 - - - ´ ´ ´ = = ´ ´ ´ Hence the correct answer will be (C) 12. A stream of electrons of velocity 3 × 107 m/s is deflected 2 mm in passing 10 cm through an electric field of 1800 V/m perpendicular to their path. The value of e/m for electrons will be : (A) 1.78 × 1011 coul/kg (B) 2 × 1011 coul/kg (C) 2.22 × 1011 coul/kg (D) 3.61 × 1011 coul/kg
• 19. Sol. 2 2 2 1 1 eE x y at 2 2 m v = = 2 2 e 2yv m Ex = ( ) ( ) 2 3 7 11 2 2 2 10 3 10 20 1800 .1 ´ ´ ´ ´ = = ´ Hence the correct answer will be (B) 13. An electron beam passes through a magnetic field of 10–3 Wb/m2 and an electric field of 3.0 × 104 V/m, both acting simultaneously. If the path of the electrons remains undeviated, then the speed of electrons will be : (A) 3 × 107 m/s (B) 7 1 10 m/ s 3 ´ (C) 3 × 105 m/s (D) 5 1 10 m/ s 3 ´ Sol. For undeviated beam of electrons eE = evB or 4 7 3 E 3.0 10 v 3 10 m/s B 10- ´ = = = ´ Hence the correct answer will be (A) 14. If the above problem, if the electric field is removed, then the radius of electron path will be : (A) 1.77 m (B) 1.77 cm (C) 1.77 mm (D) None of these Sol. If electric field is removed, then the path of the electron becomes circular 2 mv Bev r = or 31 7 19 3 mv 9.1 10 3 10 r eB 1.6 10 10 - - - ´ ´ ´ = = ´ ´ = 0.0177m = 1.77 cm Hence the correct answer will be (B) 15. In Q.3, if the length of field region is 10 cm, then on removing the magnetic field, the vertical displacement of the electron will be : (A) 2.9 m (B) 2.9 cm (C) 2.9 mm (D) None of these Sol. 2 2 2 1 1eE 1 y at 2 2m v = = or ( ) ( ) 2 19 4 2 31 7 1.6 10 3 10 .1 8 y m 273 2 9.1 10 3 10 - - ´ ´ ´ ´ = = ´ ´ ´ ´ = 2.9 cm Hence the correct answer will be (B)
• 20. 16. A charged oil drop is suspended in a uniform electric field of 300 V/cm. If the mass of the drop is 9.75 × 10–12 gm, then charge on it will be : (A) 29.67 × 10–19 Coul. (B) 67.29 × 10–19 Coul. (C) 31.85 × 10–19 Coul. (D) None of these Sol. In the stationary state of drop qE = mg or 15 2 mg 9.75 10 9.8 q E 300 10 - ´ ´ = = ´ = 31.85 × 10–19 Hence the correct answer will be (C) 17. The distance between the plates of a C.R.O. is 6.25 × 10–3 m and potential difference between them is 3.25 × 103 V. The acceleration of an electron in between the plates will be : (A) 8.9 × 1013 m/s2 (B) 8.9 × 1014 m/s2 (C) 8.9 × 1015 m/s2 (D) 8.9 × 1016 m/s2 Sol. F eE eV a m m md = = = or 19 3 31 3 1.6 10 3.125 10 a 9.1 10 6.25 10 - - - ´ ´ ´ = ´ ´ ´ = 8.9 × 1016 m/s2 Hence the correct answer will be (D) 18. In a cathode ray tube the distance between the cathode and the anode is 0.5 m and the potential difference is 50 kV. If an electron starts from rest from the cathode, then it will strike the anode with a velocity : (A) 2.66 × 108 m/s (B) 1.33 × 108 m/s (C) 0.66 × 108 m/s (D) 0.33 × 108 m/s Sol. Kinetic energy of electron 2 1 mv eV 2 = 2eV v m = or 19 3 31 2 1.6 10 50 10 v 9.1 10 - - ´ ´ ´ ´ = ´ = 1.33 × 108 m/s Hence the correct answer will be (B) 19. An electron with energy 880 eV enters a uniform magnetic field of induction 2.5 × 10–3 T. The radius of path of the circle will approximately be : (A) 4 km (B) 4 m (C) 4 cm (D) 4 mm Sol. 2 mv evB r =  mv 2mE r eB eB = = or 31 19 19 9 2 9.1 10 880 1.6 10 r 0.04m 1.6 10 2.5 10 - - - - ´ ´ ´ ´ ´ = » ´ ´ ´ = 4 cm
• 21. Hence the correct answer will be (C) 20. The potential difference between cathode and anode of a C.R. tube is 200kV and distance between them is 25 cm. The intensity of electric field inside the tube will be : (A) 8 × 105 v/m (B) 8 × 104 v/m (C) 8 × 103 v/m (D) 0 Sol. 3 5 V 200 10 E 8 10 v /m d 0.25 ´ = = = ´ Hence the correct answer will be (A) 21. In the above problem, energy gained by the electron will be : (A) 3.2 × 10–14 eV (B) 3.2 × 10–14 erg (C) 3.2 × 10–14 MeV (D) 3.2 × 10–14 Joule Sol. E = eV = 1.6 × 120–19 × 2 × 105 = 3.2 × 10–14 Joule Hence the correct answer will be (D) 22. A beam of electrons is under the effect of 1.36 × 104 V applied across two parallel plates 4 cm apart and a magnetic field of induction 2 × 10–3 T at right angles to each other. If the two fields combinigly produce no deflection in the electron beam, then the velocity of electrons will be : (A) 2.7 × 107 m/s (B) 1.7 × 107 m/s (C) 2.7 × 108 m/s (D) 1.7 × 108 m/s Sol. 4 E V 1.36 10 v B Bd 0.04 0.002 ´ = = = ´ = 1.7 × 108 m/s Hence the correct answer will be (D) 23. In the above problem if the electric field is made zero, the the radius of the path of electron will be : (A) 0.4834 m (B) 0.8434 m (C) 0.3484 m (D) 0.4304 m Sol. 31 8 19 mv 9.1 10 1.7 10 r eB 1.6 10 0.002 - - ´ ´ ´ = = ´ ´ = 0.4834 m Hence the correct answer will be (A) 24. In Q.1, if the length of field region is 20 cm, then on removing the magnetic field the deflection of electrons produced by electric field will be : (A) 4.14 km (B) 4.14 m (C) 4.14 cm (D) 4.14 mm Sol. 2 19 4 2 31 2 16 1 eE l 1.6 10 1.36 10 0.04 y 2 eB v 2 0.04 9.1 10 1.7 10 - - ´ ´ ´ ´ = = ´ ´ ´ ´ ´ = 4.14 cm Hence the correct answer will be (C)
• 22. 25. In Thomson's experiment for the measurement of e/m of an electron, the beam remains undeflected when the electric field is 105 V/m and the magnetic field is 10–2 T. The beam was originally accelerated through a potential difference of 285 volt. The value of specific charge of electron will be : (A) 1.67 × 1011 C/kg (B) 1.47 × 1011 C/kg (C) 1.74 × 1011 C/kg (D) None of these Sol. 2 1 mv ev 2 = and eE = Bev or 2 2 10 2 4 e v E 10 m 2V 2VB 2 285 10- = = = ´ ´ = 1.74 × 1011 C/kg Hence the correct answer will be (C) 26. In a Thomson set-up, electrons accelerated through 2500 V enter the region of crossed electric and magnetic fields of strengths 36 × 104 V/m and 1.2 × 10–3 T respectively and pass through undeflected. The e m of electrons will be : (A) 1.6 × 1011 C/kg (B) 1.7 × 1011 C/kg (C) 1.8 × 1011 C/kg (D) None of these Sol. 2 2 8 2 2 6 e E 3.6 10 m 2.VB 2 2500 1.2 10- ´ = = ´ ´ ´ P = 1.8 × 1011 C/kg Hence the correct answer will be (C) CMP : A gas of identical hydrogen like atoms has some atoms in ground state and some atoms in a particular excited state and there are no atoms in any other energy level. The atoms of the gas make transition to a higher energy state by absorbing monochromatic light of wavelength 340A° subsequently, the atoms emit radiation of only six different photon energies. Some of emitted photons have wavelength 340 A° some have wavelength more and some have less than 340 A° . 27. Find the principal quantum number of the initially excited state. (A) 1 (B) 2 (C) 3 (D) 4 Solution: (B) During de-excitation, photons of six different wavelengths are emitted; therefore, principal quantum number of highest excited level is 4. incidence hc E 40.8ev 340A   Emin  E4  E3 Since, some of emitted photons have energy less that 40.8 eV, Therefore Emin < 40.8 eV Also, Emax  E4 – E1, but some of emitted photons have energy greater than 40.8 eV. Therefore Emax > 40.8 eV. As the atoms of the gas make transition by absorbing photons of energy 40.8 eV. Since, atoms of ground state do not absorb the incident radiation hence the incident radiation is absorbed by initially excited atom which may belong to either n  2 or there transition is not possible.
• 23. Therefore, the only possible value of principal quantum number of initially excited level is n=2. 28. Identify the gas (Z ?). (A) 5 (B) 2 (C) 33 (D) 4 Solution: (D) The atoms of n  2 make transition to n  4 by absorbing photons of energy 40.8 eV. E4 – E2  40.8 eV But 2 n 2 13.6z E eV n  …(i) 2 2 4 2 13.6z 13.6z E ,E 16 4   Substituting these values in (i), we get 2 1 1 13.6z 40.8 Z 4 16 14            29. Find the ground state energy (in eV). (A) 13.6 eV (B) – 54.4 eV (C) – 122.4 eV (D) – 217.6 eV Solution: (D)   2 1 2 13.6 4 E 217.6eV 1    30. Find the maximum and minimum energies of emitted photons. (in eV). (A) 204, 10.6 (B) 104, 3.6 (C) 40.8, 10.6 (D) None of these Solution: (A) Emax  E4  E1 and Emin  E4  E3 Also         2 2 4 3 2 2 13.6 4 13.6 3 E 13.6eV and E 24.2eV 4 3        Emax  –13.6  217.6  204eV Emin  –13.6  24.2  10.6eV For Test 1. In a Millikan's experiment an oil drop of radius 1.5 × 10–6 m and density 890 kg/m3 is held stationary between two condenser plates 1.2 cm apart and kept at a potential difference of 2.3 kV. If upthrust due to air is ignored, then the number of excess electrons carried by the drop will be : (A) 4 × 1010 (B) 4 × 104 (C) 400 (D) 4 Sol. ne E = mg or 3 mg 4 r gd n eE 3eV p r = = = 4 Hence the correct answer will be (D)
• 24. 2. In a Millikan's set up, a charged oil drop falss under gravity with a terminal speed v. The drop is held stationary applying a suitable electric field and is found to carry 2 excess electrons. Suddenly the drop is found to carry 2 excess electrons. Suddenly the drop is found to move upwards with the same terminal velocity v. In this observation which of the following possibilities appears to be most befitting? (A) the electric field stops acting (B) the drop loses the excess electrons (C) the drop picks up some additional electrons (D) None of these Sol. Hence the correct answer will be (C) 3. In the above problem, the number of additional electrons picked up by the drop will be : (A) 2 (B) 4 (C) 8 (D) 16 Sol. In the first case 2eE = mg = 6 n rv In second case neE = mg + 6  rv = 2mg = 2 × 2eE  n = 4 Hence the correct answer will be (A) 4. A small metal plate (work function = 1.17 ev) is placed at a distance of 2 m from a monochromatic light source of wavelength 4.8 × 10–7 m and power 1.0 watt. The light falls normally on the plate. The number of photons striking the metal plate per second per unit area will be : (A) 4.82 × 1012 (B) 4.82 × 1014 (C) 4.82 × 1016 (D) 4.82 × 1018 Sol. No. of electrons striking the plate per sec. per meter2 ( ) 7 2 2 34 8 E 1 4.8 10 4 r hc 4 3.14 2 6.6 10 3 10 - - l ´ ´ = = p ´ ´ ´ ´ ´ ´ = 4.82 × 1016 Hence the correct answer will be (C) 5. In the above problem, the maximum kinetic energy of electrons emitted from the plate will be : (A) 2.253 × 10–19 J (B) 3.253 × 10–19 J (C) 5.253 × 10–19 J (D) 2.532 × 10–19 J Sol. Maximum kinetic energy of photoelectrons 34 8 max 7 he 6.6 10 3 10 E 4.8 10 - - ´ ´ ´ = - f = l ´ – 1.17 × 1.6 × 10–19 = 4.125 × 10–19 – 1.872 × 10–19 = 2.253 × 10–19 J Hence the correct answer will be (A) 6. In Q.9, the maximum speed of emitted electrons will be : (A) 7.036 × 107 m/s (B) 7.036 × 106 m/s (C) 7.036 × 105 m/s (D) 7.036 × 104 m/s