Atoms and Nuclei Atoms Notes

ATOMS AND NUCLEI ATOMS Thomson's Atomic Model : According to this model, an atom consists of a positively charged sphere in which entire mass & positive charge of the atom is uniformly distributed. Inside this sphere, the electrons are embedded like seeds in a waternelon or like plums in a pudding. The number of electrons is such that their negative charge is equal to positive charge. Thus, atoms is electrically neutral. ––– + + – – – + + – – + – Limitations of Thomson's Model : (i) Could not explain the origin of spectral series of hydrogen & other atoms. (ii) Could not explain large angle scattering of  - particles observed by Rutherford. Rutherford's -scattering experiment : An  - particle is He nucleus containing 2 protons & 2 neutrons. It has 4 units of mass & 2 units of positive charge. Many radioactive elements emit  - particles. Gold foil S Lead cavity ZnS screen  Collimator S is a radioactive source contained in a lead cavity. The  - particle emitted by the source are collimated into narrow beam with the help of collimator. The collimated beam is allowed to fall on a thin gold fail of thickness  10–5 m.  - particles are scattered in different directions are observed though a rotatable detector consisting of ZnS screen & a microscope. The  particles produce bright flashes on ZnS screen. These are observed by the microscope &counted at different scattering angle . Observations : (i) Most of the - particle pass straight through the gold foil or suffered very small angle of deflections. (ii) A few - particles scatter through large ogles ( > 90°). (iii) Rarely, an  - particle rebounces i.e., scattered through an angle of 180° Explanation : (i) Since most of the - particles passed undeviated, the atom has a lot of empty space in it. (ii) To explain large scattering of  - particles, Rutherfored suggested that all the positive charge & entire mass of the atom is confined to an extremely small central core called as nucleus. (iii) The scattering of  - particles through different angles was explained as : The particles I, I. Which pass through the atom at a large distance from from the nucleus experience a small electrostatic force of repulsion & undergo a small defection. The  particles 2. 2 which pass through the atom at a close distance from the nucleus suffer a larger defection. The  - particle 3, which travels directly towards the nucleus shows down, comes to rest & is deflected through 180° and hence retraces its path. ATOMS AND NUCLEI 1 2 3 2’ 1’ + Number of scattered particle N The graph between scattering angle  & the number of  - particles directly scatted N is as shown : 180° Scattering angle () 0° Distance of closet Approach : An  - particle travelling towards the center of the nucleus slows down as it apparoaches the nucleus. At a certain distance, say r0 from the nucleus, the -particle comes to rest for a moment and then retraces it path. It initial kinetic energy is completely converted into electrostatic potential energy. This distance r0 is called the distance of closest approach. This distance gives an estimate fo the size of the nucleus. Mathematically. +Ze h0 1 1 (2e) ( Ze ) mv 2  2 4 0 r0 where 2 e is the charge on - particle and Ze is the charge on nucleus. Ze( 2e) 1  4 0  mv 2  2  Impact Parameter : Impact parameter is defined as the perpendicular distance of the velocity vector fo the  - particle from the center of the nucleus, when it is far away from the nucleus of the atom. Rutherford derived the relation between impact parameter and scattering angle, which is given by. r0  velocity vector of  particle   b Nucleus b 1 Ze 2 cot  / 2 1 40 mv 2 2 Rutherford's Model : On the basis of  - scattering experiment, Rutherford proposed the model of an atom as : ATOMS AND NUCLEI 1. 2. 3. 4. 1. 2. 1. An atom consists of a small and massive central core in which the entire positive charge and almost the whole mass are concentrated. This core is called a smucleus. The size of the nucleus is very small as compared to the size of atom. The nucleus is surrounded by a number of electrons so that their total negative charge is equal to the total positive charge and the atom is electrically neutral. The electrons revolve around the nucleus in various orbits. The centripetal force required for their revolution is provided by the electrostatic attraction between the electrons and the nucellus. Drawbacks of Rutherford's model of Atom : It failed to explain the stability of the atom. The electrons revolving around the nucleus are accelerated towards the nucelus of the atom. Any charged particle while accelerating ore retarding loses energy through electromagnetic radiations. As a result, the radius of the path of the electron should go on decreasing and ultimately it should fall into the nucleus by following spiral path. But, in fact, it never happens. It contradicts the stability of matter. It failed to explain the spectrum of an atom. As electron can revolve around a nucleus in circular orbits of all possible radii, so atom should emit continuous energy spectrum. But even the simplest atom i.e. hydrogen atom has line spectrum instead of continuous spectrum, which Rutherford's model could not explain. Bohr's Model of Hydrogen Atom : An atom consists of a central core called nucleus, in which entire positive charge and almost entire mass of the atom is concentrated. Electron revolve around the nucleus in circular orbits. The centripetal force required for revolution is provided by the electrostatic force of attraction between he electron and the nucleus. If m is the mass of electron moving with a velocity v in a circular orbit of radius r, then the necessary centripetal force if F = mv 2 r Also, the electrostatic force of attraction between the nucleus of charge (+Ze) and electron of charge (–e) is F 2. 1 ( Ze)( e) KZe 2  4 0 r2 r2 1 where K = 4 0 mv 2 KZe 2  2 r r According to Bohr, electrons can revolve only is those certain discrete non radiating orbits, called sateionary orbits, for which total angular momentum of the revolving electron is an integral multiple of h/2, where h is Plank's constant. Thus the angular momentum of the orbiting electron is quantized. While revolving in these stationary orbits, an electron does not radiate energy. For any permitted stationary orbit. nh 2 where n is any positive integer, 1, 2, 3 ............... It is called principal quantum number. The emission/absorption of energy occurs only when an electron jumps from one stationary non-radiating orbit to another. The difference in the total energy of electron in the two permitted orbits is absorbed when the electron jumps from inner to the outer orbit i.e., absorbs photons, and emitted when electron jumps from outer to the inner orbit i.e., emits photons. If E1 is total energy of electron in an inner stationary orbit and E2 is its total energy mvr  3. in an outer stationary orbit, then hv  E 2  E1 Some quantities related to Bohr's orbit : ATOMS AND NUCLEI (a) Radii of Bohr's statioanry orbits : We know, mvr = Substituting v in the equation nh nh or v = 2 2mr mv 2 KZe 2  2 r r m n2h2 KZe2  2 2 2 2 r 4 m r r 2 2 nh r for hydrogen atom Z = 1 4 2mKZe 2 r n2h2 This shows that r  n2 i.e. radii of stationary orbits are in the ratio 12 : 22 42mKe2 : 33 and so on i.e. 1 : 4 :9 ........ Clearly the stationary orbits are not equally spaced. (b) Velocity of electron in Bohr's sationary orbit : We have, nh KZe 2 Also, r = 2m  mv 2 KZe 2 nh 2KZe 2  or v = 2 2mv mv nh r Therefore, 2Ke 2 nh Total energy of electron : The energy of electron revolving in a stationary orbit is of two For hydrogen atom Z = 1 (c) Therefore, v types : Kinetic energy is due to motion and potential energy is due to position of electron. We have. mv 2 KZe 2  2 r r 2 1 KZe 2 i.e., K.E. of electron = mv  2 2r KZe( e) KZe 2  Potential energy of electron = r r Total energy of electron in the orbit E = K.E. + P.E. = Substituting r = n 2h 2 4 mKZe 2 2 For hydrogen atom Z = 1, we get, E    E– 1 KZe 2 KZe 2 KZe 2   2 r r 2r 2 2 mK 2 Z 2 e 4 n 2h 2 22mK 2 e4 n2h2 13 .6 eV n2 Total energy of electron in a stationary orbit is negative, which means the electron is bound Substituting the standard values, we get E to the nucleus by means of electrostatic attraction. Origin of spectral lines : Suppose E1 = total energy of electron in the inner (n1 th) orbit and e2 = total energy of electron in the outer (n2th) orbit. When an electron jumps from an outer to an inner orbit, the frequency of radiation emitted, according to Bohr's third postulate is hv = E2 – E1 ATOMS AND NUCLEI hv   2 2mK 2 e 4 2 2mK 2 e 4  n 22h 2 n12h 2 hc 2 2mK 2 e 4   h2 1 1  2  2  n1 n 2  1 2 2mK 2 e 4  1 1   2  2 2  h  n1 n 2  E2 Photon h E1 + 1 =  , the wave number of radiation emitted i.e., number of complete wave in unit  length. Now, and 2 2mK 2 e 4  R is a constant called Rydberg constant. R= 1.097 × 107 m–1 ch3 1 1  From above equation   R  2  2   n1 n2  1. This equation is called the Rydberg formula for the spectrum of hydrogen atom. Bohr's explanation of Spectral of Hydrogen Atom : Lyman Series : Bohr postulated that Lyman series of obtained when an electron jumps to the first orbit (n1 = 1) from any outer orbit (n2 =2, 3, 4 ........) 1 1   R  2  2  where n : 2, 3, 4............ This series lies in the ultra violet region of 2  1 n2  2. the spectrum and agree well with the values of v observed experimentally by Lyman. Balmer Series : According to Bohr, Balmer series is obtained when an electron jumps to the second orbit (n1 = 2) from any outer orbit (n2 = 3, 4, 5 .......) Wave number of these spectral lines is 1 1   R  2  2  where n = 3, 4, 5......... This series lie in the visisble part of the 2 n2  2 3. spectrum. Paschen Series : According to Bohr Paschen series is obtained when an electron jumps to the 3 rd orbit (n1 = 3) from any out of orbit (n2 = 4, 5, 6......) Bohr calculated the wave number of spectral lines of Paschen series from the relation. 1 1   R 2  2  n2  3 where n2 = 4, 5, 6 ........ This series lies in the infrared region of the spectrum. 4. Brackett Series : According to Bohr, Brackett series is obtained when an electron jumps to the 4 th orbit (n1 = 4) from any out orbit (n2 = 5, 6, 7 ..........). This series lie in far infrared region.  1 1   R 2  2  n2  4 where n2 = 5, 6, 7. ................... ATOMS AND NUCLEI 5. Pfund Series : According to Bohr, Pfund series is obtained when an electron jumps to the 5th orbit (n1 = 5) from any outer orbit (n2 = 6, 7, 8........). This series lies in far infrared region. 1 1   R 2  2  n2  5 where n2 = 6, 7, 8 ................ Energy Level Diagram : A diagram which represents the total energies of electron in different sationary orbits of an atom are called the energy level diagram and are represented by parallel horizonal lines. Total energy in the nth orbit of hydrogen atom is given by E  On substituting the values, we get E  – 2 2mK 2 e 4 n 2h 2 13 .6 eV n2 Substituting n = 1, 2, 3............ we get the energies of electrons in various stationary orbits as : E1  13.6  13.6 eV 12 E2  13 .6  3.4 eV 22 E3  13 .6  1.51eV 32 E4  13.6  0.85 eV 42 E5  13.6  0.54 eV 52 E6  13 .6  0.37 eV 62 E7  13.6  0.28 eV 72 Clearly, as n increases, En becomes less negative until at n = , En = 0