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Chapter Contents Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Introduction Alpha-Particle Scattering and Rutherford’s Nuclear Model of atom Bohr Model of the Hydrogen Atom The Line Spectra of the Hydrogen Atom Hydrogen Spectrum X-rays and the Atomic Number Spontaneous and Stimulated Emission Maser and Laser Introduction In this chapter, we will study about various atomic model. Initially J.J. Thomson proposed an atomic model in which he thought of as electrons embedded in between protons. In 1911, his student Earnest Rutherford proposed a nuclear model, on the basis of scattering experiment. In order to rectify the short comings of Rutherford’s model, in 1913, Niels Bohr combined classical and early quantum concepts of Einstein and Plank to explain the stability of an atom. In this chapter we will confine our study to Hydrogen atom. ALPHA-PARTICLE SCATTERING AND RUTHERFORD’S NUCLEAR MODEL OF ATOM In this experiment, he used a beam of 5.5 MeV, -particles obtained from 214 83 Bi radioactive source and bombarded it on a thin gold foil. Scattering of -particles was observed through a rotatable detector made up of zinc sulphide screen and a microscope. Observations : 1. Most of the -particles passed through the foil without any deviation. 2. About 0.14% of the incident -particles scattered by more than 1°. 3. Deflection of more than 90° was observed in about 0.0125% of the incident -particles. Inferences : 1. Most of the space in an atom is unoccupied as about 99.86% - particles passed without deviation. 2. There must be an extremely small region of concentrated positive charge at the centre of an atom. This small region is called nucleus. The scattering of -particles is due to encounter between the -particle and the nucleus of the atom. Chapter 27 Atoms
72 Atoms NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 3. The nucleus of the atom is so massive as compared with the -particles that it remains at rest during the encounter, whereas electrons, owing to their little mass, cannot appreciably deflect the far more massive -particles. 4. Electrons revolve around the nucleus in orbits just like planets revolve around the sun. 5. Making these assumptions Rutherford was able to establish a formula for calculating the number of -particles scattered at an angle  as : N ()  4 2 2 . 2 sin E Z        ... [Number of -particles] where, E = K.E of -particles = 2 2 1 mv . Impact parameter The perpendicular distance of the initial velocity vector of the -particle from the centre of nucleus is called impact parameter. +2e  b Ze Fig. Schematic Diagram Showing Path of an -particle Rutherford showed that the angle of scattering () is related to the impact parameter (b) according to the equation. E Ze b 0 2 4 2 cot          The impact parameter b is the perpendicular distance of the initial velocity vector of the -particle from the centre of the nucleus. Thus, smaller the impact parameter b, greater the scattering. Distance of Closest Approach Let r 0 be the distance of closest approach of the -particle to the nucleus. The (positive) charge on the nucleus is Ze, and that on the -particles is 2e, where e is the electronic charge. At distance of closest approach, the -particles is momentarily at rest and the initial kinetic energy E is entirely converted into electrostatic potential energy. Hence, at this instant E = 4 0 1  0 2 2 r Ze  E Ze r 2 0 0 2 4 1   ...[Distance of closest approach]
NEET Atoms 73 Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Note : (1) The scattering is proportional to the target thickness for thin targets. (2) The -particle reverses its motion without ever actually touching the gold nucleus. (3) Electrons revolve around the nucleus in circular orbits so that the centripetal force is provided by the electrostatic force of attraction given by Coulomb’s law. (4) Almost all mass and total positive charge is in the nucleus of radius not more than 10–14m. (5) The size of atom is about 104 times the size of the nucleus i.e. the radius of atom is about 10–10m. (6) Nuclear collision is considered to be perfectly elastic and obey the law of conservation of energy, momentum and angular momentum. (7) Rutherford’s model is unstable due to loss of energy in the form of electromagnetic radiation by orbiting electrons. (8) According to Rutherford’s model atoms should emit continuous radiation of all frequencies which is against the observed atomic spectra. Electron Orbits Rutherford assumed atom to be an electrically neutral sphere having nucleus at the centre and electrons revolving around the nucleus in different orbits. The electrostatic force of attraction between the revolving electrons and nucleus provides the centripetal force to electrons to keep them in their circular orbits. So, for stable orbit in a Hydrogen atom, Electrostatic force of attraction = Centripetal force 2 0 1 4 ee  r = 2 mv r 2 2 0 1 1 2 42    e K mv r Potential energy of electron, 2 0 1 4    e U r So, the total energy E of the electron in a hydrogen atom is E = K + U = 22 2 84 8 00 0      ee e rr r Example 1 : What is the mass of an -particle? Solution : Alpha particle is a doubly charged Helium nucleus, so it has two neutrons and two protons. Therefore, M U 4 Example 2 : Most of the -particles passed through the foil undeviated. What can be concluded from this observation? Solution : It can be concluded that most of the space in an atom is unoccupied. Example 3 : According to Rutherford, which force was responsible for centripetal acceleration of an electron revolving around the nucleus? Solution : Electrostatic force of attraction. Example 4 : What should be the effect of impact parameter upon the deviation of -particle? Solution : As the impact parameter decreases, electrostatic force of repulsion and therefore deviation will increase.
74 Atoms NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 BOHR MODEL OF THE HYDROGEN ATOM Postulate I: An electron in an atom could revolve in certain stable orbits without emitting radiant energy. Each atom has certain definite stable orbits. Electrons can exist in these orbits. Each possible orbit has definite total energy. These stable orbits are called the stationary states of the atom. Postulate II: An electron can revolve around the nucleus in an atom only in those stable orbits whose angular momentum is the integral multiple of 2 h  (where h is Planck’s constant). Therefore, angular momentum (L) of the orbiting electron is quantised. 2 nh L   2   nh mvr where, n = 1, 2, 3 ..... Postulate III: An electron can make a transition from its stable orbit to another lower stable orbit. While doing so, a photon is emitted whose energy is equal to the energy difference between the initial and final states. Therefore, the energy of photon is given by, h = Ei – Ef where Ei and Ef are the energies of the initial and final states. (Ei > Ef ) Mathematical Analysis of Bohr’s Theory for Hydrogen and hydrogen like elements From the equations (i) & (ii) given below, we obtain various results. Force of attraction between electron and nuclues provides necessary centripetal force to electrons 2 0 1( ) 4 Ze e r = 2 mv r and ... (i) 2 nh mvr   ... (ii) (i) Velocity of electron in nth orbit: By putting the value of mvr in equation (i) from (ii), we get 2 2 0 1 4 Ze r = 2 2 nh v r         2 0 1 2           Ze v n h Velocity of electron for hydrogen and hydrogen like element in nth orbit. m/s 137 n  c Z v n ... (iii) where c is speed of light, 3 × 108 m/s. (ii) Radius of the nth orbit : From equation (iii), putting the value of v in equation (ii), we get,  2 2 0 2           n h r n Zme Radius of nth orbit for hydrogen and hydrogen like element 2 n 0.53 Å n r Z  ... (iv)