Modern Physics Atomic Structure

MODERN PHYSICS Structure of matter always been an interesting area of research for physicists. Till 20 th century it was assumed that matter consists of indivisible small tiny particles called “atoms”. But with the study and research it was found that the atom is divisible and made of other small partiicles called electron, proton and Neutron. So may physicists tried to explain the structure of atom but finally it was Neils Bohr whose explanation about the structure was well accepted. For simplicity they have taken hydrogen atom and then it can be extended to other H-like atoms too. Some of the historical modles are also explained and them drawbacks. VARIOUS MODELS FOR STRUCTURE OF ATOM Dalton's Theory Every material is composed of minute particles known as atom. Atom is indivisible i.e. it cannot be subdivided. It can neither be created nor be destroyed. All atoms of same element are identical physically as well as chemically, whereas atoms of different elements are different in properties. The atoms of different elements are made up of hydrogen atoms. (The radius of the heaviest atom is about 10 times that of hydrogen atom and its mass is about 250 times that of hydrogen). The atom is stable and electrically neutral. Thomson's Atom Model The atom as a whole is electrically neutral because the positive charge present on the atom (sphere) is equal to the negative charge of electrons present in the sphere. Atom is a positively charged sphere of radius 10 –10 m in which electrons are embedded in between. The positive charge and the whole mass of the atom is uniformly distributed throughout the sphere. Electron Positively charged matter Shortcomings of Thomson's model (i) The spectrum of atoms cannot be explained with the help of this model (ii) Scattering of –particles cannot be explained with the help of this model RUTHERFORD ATOM MODEL Rutherford experiments on scattering of  – particles by thin gold foil The experimental arrangement is shown in figure. –particles are emitted by some radioactive material (polonium), kept inside a thick lead box. A very fine beam of –particles passes through a small hole in the lead screen. This well collimated beam is then allowed to fall on a thin gold foil. While passing through the gold foil, –particles are scattered through different angles. A zinc sulphide screen was placed out the other side of the gold foil. This screen was movable, so as to receive the –particles, scattered from the gold foil at angles varying from 0 to 180°. When an –particle strikes the screen, it produces a flash of light and it is observed by the microscope. It was found that : (i) Most of the – particles went straight through the gold foil and produced flashes on the screen as if there were nothing inside gold foil. Thus the atom is hollow. (ii) Few particles collided with the atoms of the foil which have scattered or deflected through considerable large angles. Few particles even turned back towards source itself. (iii) The entire positive charge and almost whole mass of the atom is concentrated in small centre called a nucleus. (iv) The electrons could not deflected the path of a – particles i.e. electrons are very light. (v) Electrons revolve round the nucleus in circular orbits. So, Rutherford 1911, proposed a new type of model of the atom. According to this model, the positive charge of the atom, instead of being uniformly distributed throughout a sphere of atomic dimension is concentrated in a very small volume (Less than 10 –13m is diameter) at it centre. This central core, now called nucleus, is surrounded by clouds of electron makes. The entire atom electrically neutral. According to Rutherford scattering formula, the number of – particle scattered at an angle by a target are given by N   Where N0 N 0 nt(2 Ze 2 )2 1  16 (4  0 )2 r 2 (mv20 )2 sin 4  2 = number of – particles that strike the unit area of the scatter n = Number of target atom per m3 t = Thickness of target Ze = Charge on the target nucleus 2e = Charge on – particle r = Distance of the screen from target v0 = Velocity of – particles at nearer distance of approach the size of a nucleus or the distance of nearer approach is given by r0  1 (2 Ze)2 1 (2 Ze)2   4  0  1 4  0 E K where EK = K.E. of particle 2  2 mv0    MODERN PHYSICS Bohr's Atomic Model In 1913 Neils Bohr, a Danish Physicist, introduced a revolutionary concept i.e., the quantum concept to explain the stability of an atom. He made a simple but bold statement that "The old classical laws which are applicable to bigger bodies cannot be directly applied to the sub–atomic particles such as electrons or protons. Bohr incorporated the following new ideas now regarded as postulates of Bohr's theory. 1. The centripetal force required for an encircling electron is provided by the electrostatic attraction between the nucleus and the electron i.e. v = Absolute permittivity of free space = 8.85 × 10–12 C2 N–1 m–2 m = Mass of electron v = Velocity (linear) of electron r = Radius of the orbit in which electron is revolving. Z = Atomic number of hydrogen like atom. 0 2. 1  Ze  e mv2  ...(i) 4 0 r 2 r Electron r + Nucleus +Ze Electrons can revolve only in those orbits in which angular momentum of electron about nucleus is an integral multiple of h nh . i.e., mvr = ...(ii) 2 2 n = Principal quantum number of the orbit in which electron is revolving. 3. Electrons in an atom can revolve only in discrete circular orbits called stationary energy levels (shells). An electron in a shell is characterised by a definite energy, angular momentum and orbit number. While in any of these orbits, an electron does not radiate energy although it is accelerated. 4. Electrons in outer orbits have greater energy than those in inner orbits. The orbiting electron emits energy when it jumps from an outer orbit (higher energy states) to an inner orbit (lower energy states) and also absorbs energy when it jumps from an inner orbit to an outer orbit. En – Em = hn,m E3 E2 E1 where, En = Outer energy state Em = Inner energy state + n,m = Frequency of radiation 5. The energy absorbed or released is always in the form of electromagnetic radiations. Nucleus MATHEMATICAL ANALYSIS OF BOHR'S THEORY 1  Ze  e mv2 nh  From above equation (i) and (ii) i.e., and mvr = ...(ii) 4 0 r 2 r 2 We get the following results. 1. Velocity of Electron in nth Orbit : By putting the value of mvr in equation (i) from (ii) we get 1  nh  Z  e2  Z Ze 2     v  v     v0  2 4  0 n  2 0 h  n Where, v0 = 1.6  10  19 2 2  8.8 5  10 12  6.62 5  10 34 where c = 3 × 10 m/s = speed of light in vacuum 8 ....(iii) = 2.189 × 106 ms–1 = c = 2.2 × 106 m/s 137 2. Radius of the nth orbit : From equation (iii), putting the value of v in equation (ii), we get Z n 2  0 h 2  n 2 e2  nh r0 r  m  r =   Z  me 2  Z 2  n 2 0 h  where 3. r0 =  8 .85  1 0 12  6 .6 2 5  10 34 3 .14  9 .11  10 3 1   1 .6  1 0  2  19 2 = 0.529 × 10–10 m  0 .53 Å Total energy of electron in nth orbit : From equation (i) KE  1 Ze 2 1  Ze  e  mv2  and PE   2K.E.  |PE| = 2 KE 2 8  0 r 4 0 r Total energy of the system E = KE + PE = –2KE + KE = – KE = By putting the value of r from the equation (iv), we get E  where E 0  4.    9 .1 1  1 0 3 1 .6  10 19  8  8.85  10 12   me 4  Z2   8 2 h 2   n 2  E0   0   6 .625  10  2 34 2  4  8.85  10 12 9.11  10 31 ...(v)  13.6 eV Time period of revolution of electron in nth orbit : T  where, T0  6. Z2 n2  Ze 2 8  0 r 4 By putting the values of r and v, from (iii) and (iv) T  5. ...(iv)   6 .625  1 0   1.6  10  2 34 1 9 4 2 r v n 3  4 20 h 3  n 3    T0 Z2  me 4  Z2 3 = 1.51 × 10–16 second Frequency of revolution in nth orbit :   4 9.11  10 31  1 .6  10 19 1 Z2 me 4 Z2 f   3  2 3  3  f0 where, f0  2 T n 4 0 h n 4  8.85  10 12 6 .625  10 34  Wavelength of photon   3 = 6.6 × 1015 Hz me444  11 11  22 1  2 hc 1 me444  11  1 me 11  Z222  EE  hc   1    me 13.6 3.6 1 22  122 ZZ22   222  222  Z EE  EE n 2 EE n1  222 222  222  222 ZZ2  113.6 22 33  2 2 2 3 n n   88000hh nn111 nn222  88000hh cc nn111 n 222  n 111 n 222   1 1  2  2  2  Z where,  is called wave number.. n2  n1 R   R H = Rydberg constant  R =   9 .11  1 0 31  1.6  10 1 9  8  8 .8 5  10 1 2  6 .62 5  1 0  2 34  3 4  3 10  = 1.097 × 107 m–1 = 1.097 × 10–3 Å–1 (for stationary nucleus). If nucleus is not stationary (i.e., mass of nucleus is not much greater than the mass of the revolving particle like R electron), then R = where, m = mass of revolving particle and M = mass of nucleus 1m/ M MODERN PHYSICS HYDROGEN LIKE ATOMS The Bohr model of hydrogen can be extended to hydrogen like atoms, i.e., one electron atoms, the nuclear charge is +ze, where z is the atomic number, equal to the number of protons in the nucleus. The effect in the previous analysis is to replace e2 every where by ze2. Thus, the equations for, rn, vn and En are altered as under: rn = where where  0 n 2h 2 nmze 2 n2 a z 0 or rn  n2 z a0 = 0.529 Å (radius of first orbit of H) ze2 z vn = = v 2 0nh n 1 or vn  z n ....(i) ....(ii) v1= 2.19 × 106 m/s (speed of electron in first orbit of H) En = – where = mz 2 e 4 8 02n 2h 2 = z2 n2 E1 or E1 = –13.60 eV En  z2 n2 ....(iii) (energy of atom in first orbit of H) Definations Valid for Single Electron System (1) (2) (3) (4) (5) (6) Ground state : Lowest energy state of any atom or ion is called ground state of the atom. Ground state energy of H atom = –13.6 eV Ground state energy of He+ Ion = –54.4 eV Ground state energy of Li++ Ion = –122.4 eV Excited State : State of atom other than the ground state are called its excited states. n=2 first excited state n=3 second excited state n=4 third excited state n = n0 + 1 n0th excited state Ionisation energy (E.) : Minimum energy required to move an electron from ground state to n =  is called ionisation energy of the atom or ion Ionisation energy of H atom = 13.6 eV Ionisation energy of He+ Ion = 54.4 eV Ionisation energy of Li++ Ion = 122.4 eV Ionisation potential (I.P.) : Potential difference through which a free electron must be accelerated from rest such that its kinetic energy becomes equal to ionisation energy of the atom is called ionisation potential of the atom. I.P of H atom = 13.6 V I.P. of He+ Ion = 54.4 V Excitation energy : Energy required to move an electron from ground state of the atom to any other exited state of the atom is called excitation energy of that state. Energy in ground state of H atom = –13.6 eV Energy in first excited state of H-atom = –3.4 eV st excitation energy = 10.2 eV. Excitation Potential : Potential difference through which an electron must be accelerated from rest so that its kinetic energy becomes equal to excitation energy of any state is called excitation potential of that state. st excitation energy = 10.2 eV. st excitation potential = 10.2 V. (7) Binding energy or Seperation energy : Energy required to move an electron from any state to n =  is called binding energy of that state. or energy released during formation of an H-like atom/ion from n =  to some particular n is called binding energy of that state. Binding energy of ground state of H-atom = 13.6 eV Ex. First excitation potential of a hypothetical hydrogen like atom is 15 volt. Find third excitation potential of the atom. Sol. Let energy of ground state = E0 E0 = – 13.6 Z2 eV and n = 2, E2 = En = E0 n2 E0 4 given E0 – E0 = 15 4 for n = 4, E4 =  – 3E 0 = 15 4 E0 16 15 E0 15  4  15  75 – E0 = – E0 =     = 4 eV 16 16 16  3  75 third excitation potential is V 4 third exicitation energy =  SPECTRAL SERIES OF HYDROGEN ATOM It has been shown that the energy of the outer orbit is greater than the energy of the inner ones. When the Hydrogen atom is subjected to external energy, the electron jumps from lower energy state i.e. the hydrogen atom is excited. The excited state is unstable hence the electron return to its ground state in about 10 –8 sec. The excess of energy is now radiated in the form of radiations of different wavelength. The different wavelength constitute spectral series. Which are characteristic of atom emitting, then the wave length of different members of series can be found from the following relations  (i) (ii) This relation explains the complete spectrum of hydrogen. A detailed account of the important radiations are listed below. Lyman Series : The series consist of wavelength which are emitted when electron jumps from an outer orbits to the first orbit i. e., the electronic jumps to K orbit give rise to lyman series. Here n1 = 1 & n2 = 2, 3, 4, ....... The wavelengths of different members of Lyman series are : First member :In this case n1 = 1 and n2 = 2 hence   (iii) 1 1 1   R 2  2   n n  1 2  1 1  3R 1  R 2  2    4 2  1 4 4  = 1216 × 10–10 m = 1216Å 3  1 0 .9 7  10 6 3R Second member : In this case n1 = 1 and n2 = 3 hence 1 1  8R 1  R 2  2    9 3  1 9 9  = 1026 × 10–10 m =1026Å 8R 8  1 0 .9 7  1 0 6 Similarly the wavelength of the other members can be calculated.   MODERN PHYSICS (iv) Limiting members : In this case n1 = 1 and n2 = , hence 1 1  1  R 2  2   R  1    1 1  = 912 × 10–10m = 912Å R 1 0 .9 7  1 0 6 This series lies in ultraviolet region.  (v) Balmer Series : This series is consist of all wave lengths which are emitted when an electron jumps from an outer orbit to the second orbit i. e., the electron jumps to L orbit give rise to Balmer series. Here n1 = 2 and n2 = 3, 4, 5 .........The wavelength of different members of Balmer series. (vi) First member : In this case n1 = 2 and n2 = 3, hence  (vii) (viii) 36 36  = 6563 × 10–10m = 6563Å 5  1 0 .9 7  1 0 6 5R Second member :  1 1  5R 1  R 2  2    2 3   36 In this case n1 = 2 and n2 = 4, hence 1 1  3R 1  R 2  2    4  16 2 16 16  = 4861 × 10–10m = 4861Å 3R 3  1 0 .9 7  10 6 Limiting members: In this case n1 = 2 and n2 = , hence 1 1 R 4 1   = 3646Å  R 2      4 R 2 This series lies in visible and near ultraviolet region. (ix) Paschen Series : This series consist of all wavelength are emitted when an electron jumps from an outer orbit to the third orbit i. e., the electron jumps to M orbit give rise to paschen series. Here n1=3 & n2= 4, 5, 6 ..... The different wavelengths of this series can be obtained from the formula 1 1 1   R 2  2   n2  3 where n2 = 4, 5, 6 ... For the first member, the wavelengths is 18750Å. This series lies in infra–red region. (x) Bracket Series : This series is consist of all wavelengths which are emitted when an electron jumps from an outer orbits to the fourth orbit i. e., the electron jumps to N orbit give rise to Brackett series. Here n1 = 4 & n2 = 5, 6, 7, ...... 1 1 1  The different wavelengths of this series can be obtained from the formula  R  2  2   n2  4 where n2 = 5, 6, 7 .......... This series lies in infra–red region of spectrum. (xi) Pfund series : The series consist of all wavelengths which are emitted when an electron jumps from an outer orbit to the fifth orbit i. e., the electron jumps to O orbit give right to Pfund series. Here n1 = 5 and n2 = 6, 7, 8 ......... 1 1 1  The different wavelengths of this series can be obtained from the formula  R  2  2   n2  5 where n2 = 6, 7, 8 ....... This series lies in infra–red region of spectrum. The result are tabulated below S. No. Series Observed Value of n 1 Value of n2 Position in the Spectrum 1. Lyman Series 1 2,3,4...  Ultra Violet 2. Balmer Series 2 3,4,5... Visible 3. Paschen Series 3 4,5,6.... Infra–red 4. Brackett Series 4 5,6,7.... Infra–red 5. Pfund Series 5 6,7,8.... Infra–red EXCITATION AND IONISATION OF ATOMS Consider the case of simplest atom i. e., hydrogen atom, this has one electron in the innermost orbit i.e., (n = 1) and is said to be in the unexcited or normal state. If by some means, sufficient energy is supplied to the electron. It moves to higher energy states. When the atom is in a state of a high energy it is said to be excited. The process of raising or transferring the electron from lower energy state is called excitation. When by the process of excitation, the electron is completely removed from the atom. The atom is said to be ionized. Now the atom has left with a positive charge. Thus the process of raising the atom from the normal state to the ionized state is called ionisation. The process of excitation and ionisation both are absorption phenomena. The excited state is not stationary state and lasts in a very short interval of time (10–8 sec) because the electron under the attractive force of the nucleus jumps to the lower permitted orbit. This is accompanied by the emission of radiation according to BOHR'S frequency condition. MODERN PHYSICS The energy necessary to excite an atom can be supplied in a number of ways. The most commonly kinetic energy (Wholly or partly) of the electrons is transferred to the atom. The atom is now in a excited state. The minimum potential V required to accelerate the bombarding electrons to cause excitation from the ground state is called the resonance potential. The various values of potential to cause excitation of higher state called excitation potential. The potential necessary to accelerate the bombarding electrons to cause ionisation is called the ionization potential. The term critical potential is used to include the resonance, excitation and ionisation potential. We have seen that the energy required to excite the electron from first to second state is 13.6 – 3.4 = 10.2 eV from first to third state is 13.6 – 1.5 = 12.1 eV., and so on. The energy required to ionise hydrogen atom is 0 – (– 13.6) = 13.6 eV. Hence ionization potential of hydrogen atom is 13.6 volt. Successes and Limitations Bohr showed that Planck's quantum ideas were a necessary element of the atomic theory. He introduced the idea of quantized energy levels and explained the emission or absorption of radiations as being due to the transition of an electron from one level to another. As a model for even multielectron atoms, the Bohr picture is still useful. It leads to a good, simple, rational ordering of the electrons in larger atoms and quantitatively helps to predict a greater deal about chemical behavior and spectral detail. Bohr's theory is unable to explain the following facts : (i) The spectral lines of hydrogen atom are not single lines but each one is a collection of several closely spaced lines. (ii) The structure of multielectron atoms is not explained. (iii) No explanation for using the principles of quantization of angular momentum. (iv) No explanation for Zeeman effect. If a substance which gives a line emission spectrum is placed in a magnetic field, the lines of the spectrum get splitted up into a number of closely spaced lines. This phenomenon is known as Zeeman effect. Ex. A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) for this atom. Also, calculate the minimum energy (in eV) that can be emitted by this atom during de–excitation. Ground state energy of hydrogen atom is –13.6 eV. Sol. The energy released during de–excitation in hydrogen like atoms is given by : E n  E n  2 1 me 4 8 20 h 2 1 1  2  2  2 Z n n  1 2  Energy released in de–excitation will be maximum if transition takes place from nth energy level to ground state i.e., 1 1 1  2 1  2   Z = 40.8 eV E  E  13 .6   E2 n  E1  13.6  2  Z = 204 eV ...(i) & also 2 n n 2  n 1 2n 2  2n 2  2 Taking ratio of (i) to (ii), we will get 4 n  1  5  n2 =4  n=2 3 Putting n=2 in equation (i) we get Z2 =  En = – 13.6 20 4  16 Z=4 13.6  15 Z2 42 = – 217.6 eV = ground state energy 2 E1 = –13.6 × n 12 E is minimum if transition will be from 2n to 2n–1 i.e. between last two adjacent energy levels. 1  1   E min  E2 n  E2 n 1 = 13.6  2  2  4 2 = 10.57 eV 3 4   is the minimum amount of energy released during de–excitation. ...(ii) Ex. A single electron orbits around a stationary nucleus of charge +Ze where Z is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron from the second orbit to third orbit. Find (i) The value of Z. (ii) The energy required to excite the electron from the third to the fourth Bohr orbit. (iii) The wavelength of electronic radiation required to remove the electron from first Bohr orbit to infinity. (iv) Find the K.E., P.E. and angular momentum of electron in the 1st Bohr orbit. [ The ionization energy of hydrogen atom = 13.6 eV, Bohr radius = 5.3 × 10–11 m, Velocity of light = 3 × 108 m/s, Planck's constant = 6.6 × 10–34 J–s ] Sol. The energy required to excite the electron from n1 to n2 orbit revolving around the nucleus with charge +Ze is given by E n 2  E n1  1 1  me 4  1 1   2  Z2  E n 2  E n1  Z2  13 .6   2  2  2 2  2 8 0 h  n1 n 2   n1 n 2  (i) Since 47.2 eV energy is required to excite the electron from n1 = 2 to n2 =3 orbit 1  1 47 .2  36 47.2 = Z2 × 13.6  2  2   Z2 = = 24.988  25 Z=5 3  1 3.6  5 2 (ii) The energy required to excite the electron from n1 =3 to n2 =4 is given by 1  2 5  1 3.6  7 1 E4 – E3 = 13.6 Z2=5  2  2  = =16.53 eV 4  144 3 (iii) The energy required to remove the electron from the first Bohr orbit to infinity    is given by 1  1 E   E 3  1 3.6  Z2  2  2   13 .6  25 eV =340eV 1    In order to calculate the wavelength of radiation, we use Bohr's frequency relation hf    6.6  1 0 34  1 0 8  3 hc –19    36 .3 9 7 Å = 13.6 × 25 × (1.6 × 10 )J   1 3.6  2 5  1 .6  10 19 (iv) K.E. =   1 1 Ze 2 mv12   = 543.4 × 10–19 J P.E. = – 2 × K.E. = – 1086.8 × 10–19 J 2 2 4 g 0 r1 Angular momentum = mv1r1 = h =1.05 × 10–34 Js 2 The radius r1 of the first Bohr orbit is given by r1 = Ex.  0 h 2 1 0.53  10 10   5 me 2 Z  0 h 2  10 –10 Q me 2  0.5 3  10 m  = 1.106 × 10 m = 0.106 Å An isolated hydrogen atom emits a photon of 10.2 eV. (i) Determine the momentum of photon emitted (ii) Calculate the recoil momentum of the atom (iii) Find the kinetic energy of the recoil atom [Mass of proton= mp = 1.67 × 10–27 kg] MODERN PHYSICS Sol. E 10.2  1 .6  1 0 1 9  = 5.44 × 10–27 kg–m/s c 3  10 8 (ii) Applying the momentum conservation p2 = p1 = 5.44 × 10–27 kg–m/s (i) Momentum of the photon is p1= p2 p1 atom Photon 2  1 p p2 (iii) K = mv2 (v = recoil speed of atom, m = mass of hydrogen atom) K = m   =  m  2m 2 2 5 .4 4  1 0  27 2 Substituting the value of the momentum of atom, we get K = Physical quantity Radius of Bohr orbit (rn) Velocity of electron in nth Bohr 2  1 .6 7  10 27 Formula rn  vn= = 8.86 × 10–27J Ratio Formulae of hydrogen atom n2h2 n2 Å 2 ; rn = 0.53 4  mkZe Z 2 Z 2 kZe 2 ; vn=2.2 × 106 nh n r1: r2:r3...rn =1:4:9...n2 1 1 1 : ... 2 3 n v1:v2:v3...vn = 1 : orbit (vn) Momentum of electron (pn) pn= Z 2 mke 2 z ; pn  n nh p1: p2 : p3 ... pn = 1 : Angular velocity of electron(n) n= 8  3 k 2 Z2 mc 4 Z2 ;   n n3h3 n3 1:2:3...n = 1 : Time Period of electron (Tn) Tn= n3h3 n3 2 2 4 ; T n  2 4  k Z me Z T1:T2:T3...Tn = 1:8:27: ... :n3 Frequency (fn) fn = Z2 4  2 k 2 Z2 e 4 m ; f  n n3 n3h3 f1 : f2 : f3 ... fn = 1 : 1 1 1 : ... 8 2 7 n3 Orbital current (In) In = 4  2 k 2 Z2 me 5 Z2 ; In  3 3 3 n h n I1 : I2 : I3 ... In =1 : 1 1 1 : ... 3 8 27 n Angular momentum (Jn) Jn= nh ;J n 2 n J1:J2:J3...Jn = 1 : 2 : 3...n Centripetal acceleration (an) an= 1 6  4 k 3 Z3 me 6 Z3 ; an 4 4 4 n h n Kinetic energy (E k n ) E Kn  Potential energy (Un) Un= Z2 2 RchZ2 ; U  n n2 n2 Total energy (En) En= RchZ2 Z2 ; En  2 2 n n 2 a1 : a2 : a3 ... an = 1 : 1 1 1 : ... 3 8 27 n 1 1 1 : ... 4 1 6 81 n 2 Z RchZ2 ; E Kn  2 2 n n 1 1 1 : ... 2 3 n E K1 : E K2 ...E Kn = 1 : 1 1 1 : ... 2 4 9 n U1 : U2 : U3 ...Un = 1 : 1 1 1 : ... 2 n 4 9 E1 : E2 : E3 ... En = 1 : 1 1 1 : ... 4 9 n2 Ex. Sol. Calculate (a) the wavelength and (b) the frequency of the H line of the Balmer series for hydrogen. (a) H line of Balmer series corresponds to the transition from n = 4 to n = 2 level. The corresponding wavelength for H line is, 1 1   1  (1.097  10 7 )  2  2  = 0.2056 × 107  4  2 (b)  = c 3.0  108 = = 6.12 × 1014 Hz  4.9  10 7   = 4.9 × 10 –7 m Ans. Ans. Ex. Find the largest and shortest wavelengths in the Lymen series for hydrogen. In what region of the electromagnetic spectrum does each series lie? Sol. The transition equation for Lymen series is given by,  1 1 1 R 2  2  (1) n   for largest wavelength, n = 2, 3, ...... n=2 1 1 1   1.097  10 7    = 0.823 × 107  max 1 4  max = 1.2154 × 10–7 m = 1215 Å  Ans. The shortest wavelength corresponds to n =  1   max or 1 1   1.097  10 7    1   min = 0.911 × 10–7 m = 911 Å Ans. Both of these wavelengths lie in ultraviolet (UV) region of electromagnetic spectrum. Ex. How may different wavelengths may be observed in the spectrum from a hydrogen sample if the atoms are excited to states with principal quantum number n ? Sol. From the nth state, the atom may go to (n – 1)th state, ...., 2nd state or 1st state. So there are (n – 1) possible transitions starting from the nth state. The atoms reaching (n – 1)th state may make (n – 2) different transitions. Similarly for other lower states. The total number of possible transitions is (n – 1) + (n – 2) + (n – 3) +............2 + 1 = Ex. n(n  1) 2 (Remember) (a) Find the wavelength of the radiation required to excite the electron in Li ++ from the first to the third Bohr orbit. (b) How many spectral linea are observed in the emission spectrum of the above excited system? Sol. (a) The energy in the first orbit = E1 = Z2 E0 where E0 = – 13.6 eV is the energy of a hydrogen atom in ground state thus for Li++, E1 = 9E0 = 9 × (– 13.6 eV) = – 122.4 eV The energy in the third orbit is E3 = Thus, E 1 E1  = – 13.6 eV n2 9 E3 – E1 = 8 × 13.6 eV = 108.8 eV. MODERN PHYSICS Energy required to excite Li++ from the first orbit to the third orbit is given by E3 – E1 = 8 × 13.6 eV = 108.8 eV. The wavelength of radiation required to excite Li++ from the first orbit to the third orbit is given by hc  E 3  E1  or  hc E 3  E1 = 1240 eV  nm  11.4 nm 108.8 eV (b) The spectral lines emitted are due to the transitions n = 3  n = 2, n = 3  n = 1 and n = 2  n = 1. Thus, there will be three spectral lines in the spectrum. Ex. Find the kinetic energy potential energy and total energy in first and second orbit of hydrogen atom if potential energy in first orbit is taken to be zero. Sol. E1 = – 13.60 eV E2 = E1 = – 3.40 eV (2)2 K1 = – E1 = 13.60 eV U1 = 2E1 = –27.20 eV K2 = 3.40 eV U2 = – 6.80 eV and Now U1 = 0, i.e., potential energy has been increased by 27.20 eV while kinetic energy will remain unchanged. So values of kinetic energy, potential energy and total energy in first orbit are 13.60 eV, 0, 13.60 respectively and for second orbit these values are 3.40 eV, 20.40 eV and 23.80 eV. Ex. A lithium atom has three electrons, Assume the following simple picture of the atom. Two electrons move close to the nucleus making up a spherical cloud around it and the third moves outside this cloud in a circular orbit. Bohr’s model can be used for the motion of this third electron but n = 1 states are not available to it. Calculate the ionization energy of lithium in ground state using the above picture. Sol. In this picture, the third electron moves in the field of a total charge + 3e – 2e = + e. Thus, the energies are the same as that of hydrogen atoms. The lowest energy is : E2 = E1  13.6 eV = = – 3.4 eV 4 4 Thus, the ionization energy of the atom in this picture is 3.4 eV. Ex. The energy levels of a hypothetical one electron atom are shown in the figure. (a) Find the ionization potential of this atom. (b) Find the short wavelength limit of the series terminating at n = 2 (c) Find the excitation potential for the state n = 3. (d) Find wave number of the photon emitted for the transition n = 3 to n = 1. (e) What is the minimum energy that an electron will have after interacting with this atom in the ground state if the initial kinetic energy of the electron is (i) Sol. 6 eV (ii) 11 eV (a) Ionization potential = 15.6 V (b) min = 12400 = 2340 Å 5.3 (c) E31 = – 3.08 – (– 15.6) = 12.52 eV Therefore, excitation potential for state n = 3 is 12.52 volt. (d) 1  E 31  31 = 12400 Å–1 = 12.52 –1 Å  1.01 × 107 m–1 12400 (e) (i) E2 – E1 = 10.3 eV > 6 eV. Hence electron cannot excite the atoms. So, Kmin = 6 eV. (ii) E2 – E1 = 10.3 eV < 11 eV. Hence electron can excite the atoms. Ex. Sol. So, Kmin = (11 – 10.3) = 0.7 eV. A small particle of mass m moves in such a way that the potential energy U = ar2 where a is a constant and r is the distance of the particle from the origin. Assuming Bohr’s model of quantization of angular momentum and circular orbits, find the radius of nth allowed orbit. The force at a distance r is, F=– dU = – 2ar dr Suppose r be the radius of nth orbit. The necessary centripetal force is provided by the above force. Thus, mv 2 = 2ar r Further, the quantization of angular momentum gives, mvr = nh 2 Solving Eqs. (i) and (ii) for r, we get 1/4  n 2h 2  r 2   8am  Ans. Ex. An imaginary particle has a charge equal to that of an electron and mass 100 times the mass of the electron. It moves in a circular orbit around a nucleus of charge + 4e. Take the mass of the nucleus to be infinite. Assuming that the Bohr’s model is applicable to the system. (a) Derive and expression for the radius of nth Bohr orbit. (b) Find the wavelength of the radiation emitted when the particle jumps from fourth orbit to the second. Sol. (a) We have m pv2 rn  ze 2 4 0 rn2 1 .....(i) The quantization of angular momentum gives, mp vrn = nh 2 ......(ii) Solving Eqs. (i) and (ii), we get n 2 h 20 r = z m e 2 p Substituting mp = 100 m where m = mass of electron and z = 4 we get, n 2h 20 rn = 400 me 2 Ans. MODERN PHYSICS (b) As we know, Energy of hydrogen atom in ground state = – 13.60 eV  z2  m 2  n  and En   For the given particle, E4 = and E2 = (13.60)(4) 2 × 100 = –1360 eV (4) 2 (13.60) (4) 2 × 100 = – 5440 eV (2) 2 DE = E4 – E2 = 4080 eV  12400 = 3.0 Å 4080  (in Å) = Ans. Ex. A particle known as -meason, has a charge equal to that of an electron and mass 208 times the mass of the electron. It moves in a circular orbit around a nucleus of charge +3e. Take the mass of the nucleus to be infinite. Assuming that the Bohr’s model is applicable to this system, (a) derive an expression for the radius of the nth Bohr orbit, (b) find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for a hydrogen atom and (c) find the wavelength of the radiation emitted when the –meson jumps from the third orbit to the first orbit. Sol. (a) We have, mv 2 Ze 2  r 4  0 r 2 or, v 2 r  Ze 2 4  0 m The quantization rule is vr = The radius is r = ...(i) nh 2m 40 m (vr) 2 = 2 Ze2 vr = n 2 h 20 Zme 2 ....(ii) For the given system, Z = 3 and m = 208 me. Thus r  n 2 h 20 624 m e e 2 (b) From (ii), the radius of the first Bohr orbit for the hydrogen atom is rh  For r = rh, h 2 0 m e e 2 h 20 n 2 h 20 = m e e 2 624 m e e 2 or, n2 = 624 or, n = 25 (c) From (i), the kinetic energy of the atom is Ze 2 mv 2 = 8  0 r 2 and the potential energy is – The total energy is En = Using (ii), Ze 2 4  0 r Ze 2 8  0 r 9  208m e 4 1872  m e e 4  Z 2 me 4  2 2  En = – =– = 8  02 n 2 h 2 8 02 n 2 h 2 n 2  8 0 h   mee4  But   2 2  is the ground state energy of hydrogen atom and hence is equal to – 13.6 eV..  8 0 h  From (iii), En = – 1872  25459.2 eV × 13.6 eV = 2 n n2 Thus, E1 = – 25459.2 eV and E3 = E1 = – 2828.8 eV. The energy difference is E3 – E1 = 22630.4eV.. 9 The wavelength emitted is = 1240 eV  nm hc = 22630.4 eV = 55 pm. E Ex. A gas of hydrogen like atoms can absorb radiations of 68 eV. Consequently, the atoms emit radiations of only three different wavelength. All the wavelengths are equal or smaller than that of the absorbed photon. (a) Determine the initial state of the gas atoms. (b) Identify the gas atoms. (c) Find the minimum wavelength of the emitted radiations. (d) Find the ionization energy and the respective wavelength for the gas atoms. Sol. (a) n(n  1) 3 2  n=3 i.e., after excitation atom jumps to second excited state. Hence nf = 3. So ni can be 1 or 2 If ni = 1 then energy emitted is either equal to, greater than or less than the energy absorbed. Hence the emitted wavelength is either equal to, less than or greater than the absorbed wavelength. Hence ni  1. If ni = 2, then Ee  Ea. Hence e  0 (b) E3 – E2 = 68 eV   1 1 (13.6) (Z2)    = 68 4 9 Z=6 MODERN PHYSICS min = (d) Ionization energy = (13.6) (6)2 = 489.6 eV = Ex. Sol. 12400 12400 = = 28.49 435.2  1 (13.6) (6) 2  1    9 12400 = E 3  E1 (c) Ans. Ans. 12400 = 25.33 Å 489.6 Ans. An electron is orbiting in a circular orbit of radius r under the influence of a constant magnetic field of strength B. Assuming that Bohr’s postulate regarding the quantisation of angular momentum holds good for this electron, find (a) the allowed values of the radius ‘r’ of the orbit. (b) the kinetic energy of the electron in orbit (c) The potential energy of interaction between the magnetic moment of the orbital current due to the electron moving in its orbit and the magnetic field B. (d) The total energy of the allowed energy levels. (a) radius of circular path r= mv Be mvr = ....(i) nh 2 ....(ii) Solving these two equations, we get nh 2Be r= (b) K = nhBe 1 mv2 = 2 4m e T (c) M = iA =   (r2) = Ans. evr e = 2 2 nh 2Be Now potential energy U = – M . B = (d) E = U + K = nhBe 2m 2 and v = nhe nhBe = 2m 2 4m nheB 4 m nheB 2 m EFFECT OF NUCLEUS MOTION ON ENERGY OF ATOM Let both the nucleus of mass M, charge Ze and electron of mass m, and charge e revolve about their centre of mass (CM) with same angular velocity () but different linear speeds. Let r1 and r2 be the distance of CM from nucleus and electron. Their angular velocity should be same then only their separation will remain unchanged in an energy level. Let r be the distance between the nucleus and the electron. Then Mr1 = mr2 r1 + r 2 = r  r1 = mr Mm and r2 = Mr Mm M r2 r1 CM m Centripetal force to the electron is provided by the electrostatic force. 1 Ze 2 mr22 = 4 0 r2 So, 1 Ze 2  Mr  2  = .  r2 4 0  M m or m or r32 = 2  Mm  3 2 Ze   r  = 4  M m 0 or e2 4 0 Mm = Mm Moment of inertia of atom about CM, where  Mm  2  r = r2  M m  = Mr12 + mr22 =  According to Bohr’s theory, nh =  2 or r2 = nh 2 Solving above equations for r, we get r= 0 n 2 h 2 e 2 Z and r = (0.529 Å) n n  n  Further electrical potential energy of the system, U=  Ze 2 4  0 r and kinetic energy, U=  Z 2 e 4 4  20 n 2 h 2 1 2 1 2 2 1  = r  and K= v2 2 2 2 v-speed of electron with respect to nucleus. (v = r) K= here Ze 2  = 4  0 r 3  Z 2 e 4 Ze 2 K= = 8  0 r 8  02 n 2 h 2 2  Total energy of the system En = K + U  En = – e 4 8 0 2 n 2 h 2 this expression can also be written as En = – (13.6 eV) Z2 n2 μ   m The expression for En without considering the motion of proton is En = – considering the motion of nucleus. me 4 , i.e., m is replaced by  while 8 0 2 n 2 h 2 MODERN PHYSICS Ex. A positronium ‘atom’ is a system that consists of a positron and an electron that orbit each other. Compare the wavelength of the spectral lines of positronium with those of ordinary hydrogen. Sol. Here the two particle have the same mass m, so the reduced mass is = mM m m2 = = m  M 2m 2 where m is the electron mass. We know that En  m E 'n  1   En m 2  Hence energy of each level is halved.  Their difference will also be halved. ’n = 2n ATOMIC COLLISION In such collisions assume that the loss in the kinetic energy of system is possible only if it can excite or ionise. neutron K, v Ex. H atom at rest in ground state and free to move head on collision What will be the type of collision, if K = 14eV, 20.4 eV, 22 eV, 24.18 eV (elastic/inelastic/perectly inelastic) Sol. Loss in energy (E) during the collision will be used to excite the atom or electron from one level to another. According to quantum Mechanics, for hydrogen atom. E = {0, 10.2 eV, 12.09 eV, ........., 13.6 eV) According to Newtonion mechanics minimum loss = 0. (elastic collsion) for maximum loss collision will be perfectly inelastic if neutron collides perfectly inelastically then, Applying momentum conservation m0 = 2mf vf  Final v0 2 1 m  02 1 K  02 2 K.E. = × 2m × = = 2 2 2 4 maximum loss = K 2 According to classical mechanics (E) = [0, (a) m K ] 2 If K = 14 eV, According to quantum mechanics (E) = {0, 10.2eV, 12.09 eV} According to classical mechanics E = [0, 7 eV] loss = 0, hence it is elastic collision speed of particle changes. m vf (b) If K = 20.4 eV According to classical mechanics loss = [0, 10.2 eV] According to quantum mechanics loss = {0, 10.2eV, 12.09eV,.........} (c) (d) loss = 0 elastic collision. loss = 10.2eV perfectly inelastic collision If K = 22 eV Classical mechanics E =[0, 11] Quantum mechanics E = {0, 10.2eV, 12.09eV, ........} loss = 0 elastic collision loss = 10.2 eV inelastic collsion If K = 24.18 eV According to classical mechanics E =[0, 12.09eV] According to quantum mechanics E = {0, 10.2eV, 12.09eV, ...... 13.6eV} loss = 0 elastic collision loss = 10.2 eV inelastic collision loss = 12.09 eV perfectly inelastic collision Ex. A He+ ion is at rest and is in ground state. A neutron with initial kinetic energy K collides head on with the He + ion. Find minimum value of K so that there can be an inelastic collision between these two particle. Sol. Here the loss during the collision can only be used to excite the atoms or electrons. So according to quantum mechanics loss = {0, 40.8eV, 48.3eV, ......, 54.4eV} ....(1) Z2 En = – 13.6 2 eV n m n Now according to newtonion mechanics 4m K Minimum loss = 0 maximum loss will be for perfectly inelastic collision. let v0 be the initial speed of neutron and vf be the final common speed. so by momentum conservation mv0 = mvf + 4mvf vf = v0 5 where m = mass of Neutron  mass of He+ ion = 4m So final kinetic energy of system K.E. = 1 1 K 1 1 v2 1 2 m v 2f + 4m v 2f = .(5m). 0 = .( mv 0 ) = 2 2 5 2 5 2 25 He + MODERN PHYSICS maximum loss = K –   So loss will be 0, K 4K = 5 5 4K  5  ....(2) For inelastic collision there should be at least one common value other than zero in set (1) and (2)  4K > 40.8 eV 5 K > 51 eV minimum value of K = 51 eV. Ex. Sol. A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before collision both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state. Let K be the kinetic energy of the moving hydrogen atom and K’, the kinetic energy of combined mass after collision. From conservation of linear momentum, 2Km = 2K'(2m) p = p’ or or From conservation of energy, K = 2K’ K = K’ + E ....(i) ....(ii) K 2 Now minimum value of E for hydrogen atom is 10.2 eV. or E  10.2 eV Solving Eqs. (i) and (ii), we get E = K  10.2 2  K  20.4 eV Therefore, the minimum kinetic energy of moving hydrogen is 20.4 eV  Ex. Sol. Ans. A neutron moving with speed v makes a head-on collision with a hydrogen atom in ground state kept at rest. Find the minimum kinetic energy of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron = mass of hydrogen = 1.67 × 10–27 kg. Suppose the neutron and the hydrogen atom move at speed v1 and v2 after the collision. The collision will be inelastic if a part of the kinetic energy is used to excite the atom. Suppose an energy E is used in this way. Using conservation of linear momentum and energy. mv = mv1 + mv2 ....(i) 1 1 1 and mv2 = mv12 + mv22 + E ....(ii) 2 2 2 2 2 2 From (i), v = v1 + v2 + 2v1v2 , From (ii), v2 = v12 + v22 + Thus, 2v1v2 = 2E m 2E m Hence, (v1 – v2)2 – 4v1v2 = v2 – 4E m As v1 – v2 must be real, 4E 0 m v2 – 1 2 mv > 2E. 2 The minimum energy that can be absorbed by the hydrogen atom in ground state to go in an excited state is 10.2 eV. Thus, the minimum kinetic energy of the neutron needed for an inelastic collision is or, 1 mv 2min  2  10.2 eV  20.4 eV 2 Ex. How many head-on, elastic collisions must a neutron have with deuterium nucleus to reduce its energy from 1 MeV to 0.025 eV. Sol. Let mass of neutron = m and mass of deuterium = 2m initial kinetic energy of neutron = K0 Let after first collision kinetic energy of neutron and deuterium be K1 and K2. Using C.O.L.M. along direction of motion 2mK 0 = 2mK1 + 4mK 2 velocity of seperation = velocity of approach 4mK 2 – 2m 2mK1 = m 2mK 0 m Solving equaiton (i) and (ii) we get K1 = K0 9 Loss in kinetic eneryg after first collision K1 = K0 – K1 K1 = 8 K 9 0 ....... (1) After second collision K2 = 8 8 K K1 = . 0 9 9 9  Total energy loss K = K1 + K2 + ..... + Kn As, K = 8 8 8 K0 + 2 K 0 + .......... + n K 0 9 9 9 MODERN PHYSICS K = 1 8 1 K0 (1 + + ......... + n 1 ) 9 9 9 1  1 K 8  9 n   K0 9  1 1 9    1  =1– n 9   Here, K0 = 106 eV, K = (106 – 0.025) eV  K 0  K 0.025 1 = n = K0 9 106 Taking log both sides and solving, we get or 9n = 4 × 107 n=8 Ex. A neutron with an energy of 4.6 MeV collides with protons and is retarded. Assuming that upon each collision neutron is deflected by 45º find the number of collisions which will reduce its energy to 0.23 eV. Sol. Mass of neutron  mass of proton = m m Neutron m K0  Proton From conservation of momentum in y-direction 2mK1 sin 45º 2mK 0 – In x-direction = 2mK1 cos 45º = 2mK 2 sin  2mK 2 cos  ....(i) ....(ii) Squaring and adding equation (i) and (ii), we have 2K0 K1 K 2 = K1 + K 0 – ....(iii) From conservation of energy K 2 = K 0 – K1 ....(iv) Solving equations (iii) and (iv), we get K1 = K0 2 i.e., after each collision energy remains half. Therefore, after n collisions, 1  2 n Kn = K 0    1 0.23 = (4.6 × 10 )    2 n 6 Taking log and solving, we get n  24 Ans. 2n  4.6  10 6 0.23 Calculation of Recoil Speed of Atom on Emission of a Photon Momentum of Photon = mc = h  (a) (b) m - mass of atom According to momentum conservation mv = ' .... (i) According to energy conservation 1 hc m 2  = 10.2 eV 2 ' Since mass of atom is very large than photon hence 1 m 2 can be neglected 2 hc = 10.2 eV ' m = 10.2 eV c =   recoil speed of atom = = 10.2 cm 10.2 eV c 10.2 cm