Structure of Atom - Chemistry PDF

2 Structure of Atom 2.1 Introduction Atom is a Greek word and its...

2 Structure of Atom 2.1 Introduction Atom is a Greek word and its meaning indivisible i.e. an ultimate particles which cannot be further subdivided. John Dalton (1803 - 1808) considered that " all matter was composed of small particle called atom”. Dalton’s Atomic Theory : This theory is based on law of mass conservation and law of definite proportions. The salient feature's of this theory are :- (1) Each element is composed of extremely small particles called atoms. (2) Atoms of a particular element are like but differ from atoms of other element. (3) Atom of each element is an ultimate particle and it has a characteristic mass but is structureless. Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom (4) Atoms are indestructible i.e. they can neither be created nor be destroyed. (5) Atom of elements take part in chemical reaction to form molecule. 2.2 Sub Atomic Particles 2.2.1 Cathode Rays (Discovery of electron) In 1859, Julius plucker started the study of conduction of electricity through gases at low pressure in a discharge tube. When a high voltage of the order of 10,000 volts or more was impressed across the electrodes, some sort of invisible rays moved from the (–)ve electrode to the (+)ve electrode. Since the (–)ve electrode is referred to as cathode, these rays were called cathode rays. Found. JEE/NEET : Class IX/X v > 10,000 volts Gas at low pressure (10–4 mm of Hg) Discharge Tube Green Glow – ------ + Cathode Rays Cathode Anode To vacuum Pump – + High Voltage Generator Properties of Cathode rays (1) Like ordinary light, cathode rays come out at right angles to the surface of cathode and travel in a straight line normal to the cathode. It casts sharp shadows, if any opaque solid object is placed in their path. (2) They rotate a light paddle wheel if placed in their path and thus shows mechanical effect i.e. they have momentum and energy and therefore, mass. The cathode rays are material particles Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom (3) These rays are deflected by the electric and magnetic fields. When the rays are passed between two electrically charged plates, these are deflected towards the positively charged plate. It shows that cathode rays carry (–)ve charge. (4) Their velocities change with the applied voltage. (5) They penetrate through the metals and other substances and also produce fluorescence. (6) These rays produce heat energy when they collide with the matter. It shows that cathode rays possess Kinetic energy which is converted into heat energy when stopped by matter. (7) Cathode rays on striking solid objects of high melting point produce X-rays. (8) They affect the photographic plate. (9) They can ionise gases. (10) Their e/m is constant.   Chemistry The Electron : (i) Electron was discovered in cathode ray experiment by J.J. Thomson in 1887. (ii) The name electron was proposed by Stoney for the fundamental unit of electricity. (iii) The charge on an electron (–1.602 × 10–19 coulombs or 4.803 × 10-10 esu) was determined by Millikan in his oil drop experiment. (iv) Actual mass of an electron 9.1×10-31 kg (9.11 × 10–28g) was calculated by J.J Thomson. Out of the three fundamental particles of an atom, electron is lightest. (v) The specific charge (e/m ratio) of electrons (cathode rays) was determined by Thomson as 1.76 × 108 coulombs/gram. The specific charge of electron decreases with increase in its velocity because increase in velocity increases the mass of electron. (vi) Density of electron is found to be 2.17 × 1017 g/cc (vii) Mass of one mole of electrons (6.023 × 1023 electron) is nearly 0.55 mg. (viii) Charge on one mole of electrons is  96500 coulombs or 1 Faraday 2.2.2 Anode Rays or Positive Rays (Discovery of Proton) ⚫ The first experiment that lead to the discovery of Zns coating the (+)ive particle was conducted by 'Goldstein'.  ⚫ He used a perforated cathode in the modified cathode ray tube. ⚫ It was observed that when a high potential perforated H2 gas inside difference was applied between the electrodes, cathode at low pressure not only cathode rays were produced but also a Production of Anode rays or Positive rays new type of rays were produced simultaneously Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom moving from anode to cathode. In actual it does not produce from anode but it looks like to be produced from anode and passed through the holes or canals of the cathode. These rays were termed canal rays since these rays passed through the canals of the cathode. These were also named anode rays as these moves from anode to cathode. ⚫ When the properties of these rays were studied by Goldstein, he observed that these rays consists of positively charged particles and named them as positive rays. ⚫ The characteristics of the positive rays are as follows. (i) These rays travel in straight lines and cast a shadow of the object placed in their path. (ii) Like cathode rays, these rays also rotate the wheel placed in their path and also have heating effect. Thus, the rays possess kinetic energy i.e. mass particles are present. (iii) These rays are deflected by electric and magnetic fields towards the negatively charged plate showing thereby that these rays carry (+)ive charge. Found. JEE/NEET : Class IX/X (iv) These rays produce flashes of light on ZnS screen (v) These rays can pass through thin metal foil. (vi) These rays can produce ionisation in gases. (vii) Positive particles in these rays have e/m value much smaller than that of e-. For a small value of e/m, it is definite that positive particles possess high mass. ⚫ Accurate measurements of the charge and the mass of the particles in the discharge tube containing hydrogen, the lightest of all gases, were made by J.J. Thomson in 1906. These particles were found to have the e/m value as +9.579  104 coulomb/gram.This was the maximum value of e/m observed for any (+)ive particle. ⚫ It was thus assumed that the positive particle given by the hydrogen represents a fundamental particle of (+)ive charge. This particle was named proton by Rutherford in 1911. Its charge was found to be equal in magnitude but opposite in sign to that of electron. Thus Charge on proton = +1.602  10-19 columb i.e. one unit (+)ive charge 2.2.3 Neutron In 1920, Rutherford suggested that in an atom, there must be present at least a third type of fundamental particles which should be electrically neutral and posses mass nearly equal to that of proton. He proposed the name for such fundamental particles as neutron. In 1932, Chadwick bombarded beryllium with a stream of -particles. He observed that penetrating radiations were produced which were not affected by electric & magnetic fields. Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom These radiations consist of neutral particles, which were called neutrons. The nuclear reaction can be shown as -Particle Be -atom C -atom Neutron Charge = +2 Atomic No. =4 Atomic No. = 6 Charge = 0 Mass = 4 amu Mass = 9 amu Mass = 12 amu Mass = 1 amu 4 9 12 1 He + Be C + n 2 4 6 0 Thus a neutron is a sub atomic particle which has a mass 1.675  10-24 g approximately 1amu, or nearly equal to the mass of proton or hydrogen atom and carrying no electrical charge. ⚫ The specific charge ( e/m value) of a neutron is zero.   Chemistry POINTS TO REMEMBER ⚫ Particles carrying negative charge were called negatrons by Thomson. The name negatron was changed to 'electron' by Stoney. The specific charge of electron was found to be independent of the nature of gas and electrode used which points out that electrons are present in all atoms. 1 ⚫ Mass of electron is times that of proton. 1837 ⚫ e/m value is dependent on the nature of the gas taken in the discharge tube, i.e. (+)ive particles are different in different gases. The mass of the proton, thus can be calculated. e 1.60210−19 ⚫ Mass of the proton = = = 1.672  10-24 g = 1.672  10-27 kg e / m 9.579  104 1.672  10−24 Mass of proton (in amu) = = 1.0072 amu. 1.6610−24 CHECK YOUR LEARNING-1 Sub Atomic Particles 1. Which of the following condition is suitable for production of cathode rays in discharge tube :– (1) Low pressure, high voltage (2) Low pressure, low voltage (3) High pressure, low voltage (4) High pressure, high voltage 2. Which of the following statements is incorrect for anode rays? Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom (1) They are deflected by electric and magnetic fields. (2) Their e/m ratio depends on the gas in the discharge tube used to produce the anode rays. (3) The e/m ratio of anode ray is constant. (4) They are produced by the ionisation of the gas in the discharge tube. 3. Which of the following pairs have identical value of e/m? (1) A proton and a neutron (2) A proton and deuterium (3) Deuterium and an –particle (4) An electron and -rays. 4. The element having no neutron in the nucleus of its atom is (1) Hydrogen (2) Nitrogen (3) Helium (4) Boron 5. The study of cathode rays (i.e. electronic discharge through gases) shows that – (1) Alpha particles are heavier than protons (2) All forms of matter contain electrons (3) All nuclei contain protons (4) e/m ratio of cathode rays is constant Found. JEE/NEET : Class IX/X 6. Proton is - (1) Nucleus of deuterium (2) Ionised hydrogen molecule (3) Ionised hydrogen atom (4) An -particle 7. Which is not deflected by magnetic field ? (1) Neutron (2) Positron (3) Proton (4) Electron 8. The ratio of the "e/m" (specific charge) values of a electron and an -particle is - (1) 2 : 1 (2) 1 : 1 (3) 1 : 2 (4) None of these 9. Electron carries which charge (1) Positive (2) Neutral (3) Negative (4) None 10. The e/m ratio for cathode rays - (1) Is constant (2) Varies as the atomic number of the element forming cathode in the discharge tube changes. (3) Varies as atomic number of the gas in the discharge tube varies. (4) Has the smallest value when the discharge tube is filled with hydrogen. 2.3 Atomic Models 2.3.1 Thomson's Model of Atom Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom ⚫ Thomson was the first to propose a detailed model of e an atom. ⚫ Thomson proposed that an atom consists of a uniform sphere of sphere of positive charge in which the electrons are (+)ive charge distributed more or less uniformly. ⚫ This model of atom is known as "Plum-Pudding model" or "Raisin Pudding Model" or "Water Melon Model". Drawbacks : ⚫ An important drawback of this model is that the mass of the atoms is considered to be evenly spread over that atom. ⚫ It is a static model. It does not reflect the movement of electron. ⚫ It did not have any experimental support.   Chemistry 2.3.2 Rutherford's Scattering Experiment -scattering experiment thin gold foil (.00004 cm) most of –particles strike here (-Ray) Zns Screen Source(Ra) of slit system -rays (lead plate) =[2He4]+2 circular [doubly ionised He particle] fluorescent screen Rutherford observed that - (i) Most of the -particles (nearly 99.9%) went straight without suffering any deflection. (ii) A few of them got deflected through small angles. (iii) A very few (about one in 20,000) did not pass through the foil at all but suffered large deflections (more than 90°) or even came back in the direction from which they have come i.e. a deflection of approx 180°. Following conclusions were drawn from the above observations - (1) Since most of the -particle went straight Nucleus through the metal foil  Un deflected undeflected, it means that -particles   -particals Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom there must be very large  empty space within the  atom. (2) Since few of the largely deflected -particles Slightly deflected -particles -particles were deflected from their original paths through moderate angles; it was concluded that whole of the (+)ive charge is concentrated and the space occupied by this positive charge is very small in the atom. ⚫ When -particles come closer to this point, they suffer a force of repulsion and deviate from their paths. ⚫ The positively charged heavy mass which occupies only a small volume in an atom is called nucleus. It is supposed to be present at the centre of the atom. (3) A very few of the -particles suffered strong deflections on even returned on their path indicating that the nucleus is rigid and -particles recoil due to direct collision with the heavy positively charged mass. Found. JEE/NEET : Class IX/X Applications of Rutherford Model On the basis of scattering experiments, Rutherford proposed model of the atom, which is known as nuclear atomic model. According to this model - (i) An atom consists of a heavy positively charged nucleus where all the protons and neutrons are present. Protons & neutrons are collectively referred to as nucleons. Almost whole of the mass of the atom is contributed by these nucleon. The magnitude of the (+)ive charge on the nucleus is different for different atoms. (ii) The volume of the nucleus is very small and is only a minute fraction of the total volume of the atom. Nucleus has a diameter of the order of 10–12 to 10–13 cm and the atom has a diameter of the order of 10–8 cm. DA Diameter of the atom 10−8 = = −13 = 105, DA = 105 DN DN Diameter of the nucleus 10 Thus diameter (size) of the atom is 105 times the diameter of the nucleus. ⚫ The radius of a nucleus is proportional to the cube root of the number of nucleons within it. R  A1/3  R = R0A1/3 cm Where R0 = 1.33  10-13(a constant) and A = mass number (p + n) and R = radius of the nucleus. R = 1.33 ×10−13 × A1/3 cm Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom (iii) There is an empty space around the nucleus called extra nuclear part. In this part electrons are present. The number of electrons in an atom is always equal to number of protons present in the nucleus. As the nuclear part of atom is responsible for the mass of the atom, the extra nuclear part is responsible for its volume. The volume of the atom is about 1015 times the volume of the nucleus. Volume of the atom (10−8 )3 = −13 3 =1015 Volume of the nucleus (10 ) (iv) Electrons revolve around the nucleus in closed orbits with high speeds. The centrifugal force acting on the revolving e- is being counter balanced by the force of attraction between the electrons and the nucleus. ⚫ This model was similar to the solar system, the nucleus representing the sun and revolving electrons as planets.   Chemistry Drawbacks of Rutherford model - (1) This theory could not explain stability of atom. According to Nucleus Maxwell electron loses it energy continuously in the form of electromagnetic radiations. As a result of this, the e- should loss energy at every turn and move closer and closer to the e– nucleus following a spiral path. The ultimate result will be that it will fall into the nucleus, there by making the atom unstable. (2) If the electrons loss energy continuously, the observed spectrum should be continuous but the actual observed spectrum consists of well defined lines of definite frequencies. Hence, the loss of energy by electron is not continuous in an atom. POINTS TO REMEMBER ⚫ Thomson's model of atom explained electrical neutrality of atom. ⚫ Density of nucleus remains same irrespective of nature of element. CHECK YOUR LEARNING-2 Atomic Models 1. In rutherford's alpha ray scattering experiment, the alpha particles are detected using a screen coated with:- Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom (1) Carbon black (2) Platinum black (3) Zinc sulphide (4) Teflon 2. Which of the following was a drawback of thomson's model ? (1) It is a static model (2) It did not have any experimental support. (3) Mass of the atom is considered to be evenly spread over the atom. (4) All of these 3. Plum pudding model was given by ____ and nuclear model was given by ____ (1) Bohr, Lavoisier (2) Thomson, Rutherford (3) Thomson, Goldstein (4) Goldstein, Bohr 4 Rutherford’s experiment on scattering of alpha particles showed for the first time that atom has - (1) Electrons (2) Protons (3) Nucleus (4) Neutrons 5 -particles are represented by – (1) Lithium atoms (2) Helium nuclei (3) Hydrogen nucleus (4) None of these Found. JEE/NEET : Class IX/X 2.4 Atomic Number and Mass Number 2.4.1 Mass Number It is represented by capital A. The sum of number of Neutrons and protons is called the mass Number of the element. i.e. A = Number of protons + Number of neutrons 2.4.2 Atomic Number (z) It is represented by Z. The number of protons present in the Nucleus is called atomic number of an element. It is also known as nuclear charge. For neutral atom : Number of proton = Number of electron For charged atom : Number of e– = Z – (charge on atom), Z= number of protons only Ex. 17Cl35 → n = 18, p = 17, e = 17 Two different elements can not have the same atomic number Number of Neutrons = Mass number – Atomic number = A – Z = (p + n) – p = n Representation of element → ZXA (where X→ symbol of element) 2.4.3 Isotopes Given by Soddy, are the atoms of a given element which have the same atomic number but different mass number i.e. They have same Nuclear charge but different number of Neutrons. Ex.1 17Cl35 17Cl37 Ex.2 6C12 6C13 6C14 n = 18 n = 20 e=6 e=6 e=6 e = 17 e = 17 p=6 p=6 p=6 p = 17 p = 17 n=6 n=7 n=8 Ex.3 (Proteium Deuterium Tritium) 1H1 1H2 1H3 Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom e=1 e=1 e=1 p=1 p=1 p=1 n=0 n=1 n=2 1H1 is the only normal hydrogen which have n = 0 i.e. no neutrons Deutrium is also called as heavy hydrogen. It represent by D 2.4.4 Isobars Given by Aston, isobars are the atoms of different element which have the same mass number but different Atomic number i.e they have different number of electron, protons & neutrons But sum of number of neutrons & protons remains same. Ex.1 1H3 2He3 Ex.2 19 K40 20 Ca40 p=1 p=2 p = 19 p = 20 e=1 e=2 n = 21 n = 20 n=2 n=1 e = 19 e = 20 p+n=3 p+n=3 n+p = 40 n + p = 40   Chemistry 2.4.5 Isodiaphers They are the atoms of different element which have the same difference of the number of neutrons & protons. Ex1 5 B 11 6 C13 Ex.2 7 N15 9 F 19 p=5 p=6 p=7 p=9 n=6 n=7 n=8 n = 10 e=5 e=6 e=7 e=9 n – p =1 n – p =1 n – p =1 n – p =1 2.4.6 Isotones/Isoneutronic Species/Isotonic They are the atoms of different element which have the same number of neutrons. Ex1 1H3 2He4 Ex. 2 19 K 39 20 Ca40 p=1 p=2 e = 19 e = 20 n=2 n=2 p = 19 p = 20 e=1 e=2 n = 20 n = 20 2.4.7 Isosters They are the molecules which have the same number of atoms & electrons. Ex.1 CO2 N2O Atoms = (1 + 2) Atoms =2+1 =3 =3 Electrons = 6 + 8 × 2 Electron =7×2+8 Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom = 22 e– = 22e– Ex.2 CaO KF Atoms 2 2 Electrons 20 + 8 19 + 9 28 e– 28 e– 2.4.8 Isoelectronic Species They are the atoms, molecules or ions which have the same number of electrons. Ex.1 Cl– Ar Ex.2 H2O NH3 Electron 18 e– 18 e– Electron = 2+8 7+3 10 e– 10 e– Ex.3 BF3 SO2 Electron = 5+9×3 16+8×2 = 5 + 27 16 + 16 = 32 e– 32 e– Found. JEE/NEET : Class IX/X POINTS TO REMEMBER ⚫ Isotopes have same chemical property but different physical property. ⚫ Isotopes and isobars do not have the same value of e/m. ⚫ Isobars do not have the same chemical & physical property. ⚫ For isotones, A1– Z1 = A2– Z2 ⚫ For isodiaphers, A1– 2Z1 = A2– 2Z2 Examples Example 1: If the mass of neutrons is doubled & mass of electron is halved then find out the new atomic mass of 6C12 and the percent by which it is increased. Solution : Step-1 6C12 → e=6 p = 6 = 6amu  n = 6 = 6amu  = 12 amu   If the mass of neutrons is doubled and mass of e– is halved then. n = 12 amu  p = 6 amu  = 18 amu   Note : mass of e is negligible, so it is not considered in atomic mass. – Final mass − Initial mass 18 − 12 Step-2 % Increment =  100 =  100 = 50% Initial mass 12 Example 2: If mass of neutron is doubled, mass of proton is halved and mass of electron is doubled then find out the new atomic weight of 6C12. p = 6 amu  Solution : Step-1 6 C12 →   = 12 amu n = 6 amu  If mass of neutron is doubled, mass of proton is halved and mass of electron is Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom doubled,then new atomic mass will be : n = 12 amu  p = 3 amu  = 15 amu   Final mass − Initial mass 15 − 12 Step-2 % Increment=  100 =  100 = 25% Initial mass 12 CHECK YOUR LEARNING-3 Isotopes, Isobars and Isotones 1. The atom A, B, C are related as A → [Z(90) + n(146)], B → [Z(92) + n(146)], C → [Z(90) + n(148)], So that :- (a) A and C - Isotones (b) A and C - Isotopes (c) A and B - Isobars (d) B and C - Isobars (e) B and C - Isotopes The wrong statements are:- (1) a, b only (2) c, d, e only (3) a, c, d only (4) a, c, e only   Chemistry 2. Two nuclides A and B are isoneutronic. Their mass numbers are 76 and 77 respectively. If the atomic number of A is 32, then the atomic number of B will be :- (1) 33 (2) 34 (3) 32 (4) 30 3. What will be the difference in the mass number if the number of neutrons is halved and the number of electrons is doubled in 8O16 :- (1) 25% decrease (2) 50% increase (3) 150% increase (4) No difference 4. The isoelectronic pair of 32 electrons is :- (1) BO33– and CO32– (2) N2 and CO (3) PO43– and CO32– (4) All of the above 5. Which of the following is iso-electronic with neon? (1) O2– (2) Mg+ (3) Mg (4) Na 2.5 Developments Leading to The Bohr's Model of Atoms Two developments played a major role in the formulation of Bohr's model of atom. These were (a) Dual nature of electromagnetic radiation (b) Atomic spectra which can be explained by assuming quantized electronic energy levels in atoms. Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom 2.5.1 Electromagnetic Waves (EM waves) Theory or Radiant Energy According to this theory the energy transmitted from one body to another in the form of waves and these waves travel in the space with the same speed as light (3 × 108 m/s) and these waves are known as electro magnetic waves or radiant energy. The radiant energy do not need any medium for propagation. Ex : Radio waves, micro waves, Infra-red rays, visible rays, ultraviolet rays, x–rays, gama rays and cosmic rays.  The radiant Energy have electric and magnetic fields and travel at right angle to these fields. The upper most point of the wave is called crest and the lower most portion is called trough. Some of the terms employed in dealing with the waves are described below. Found. JEE/NEET : Class IX/X Crest Wavelength Electric field component Direction of propagation Magnetic field Trough component (1) Wavelength () (Lambda) : It is defined as the distance between two nearest crest or trough. It is measured in terms of a Å (Angstrom), pm (Picometer), nm (nanometer), cm(centimeter), m (meter). 1Å = 10–10 m, 1 pm = 10–12 m, 1nm = 10–9 m, 1cm = 10–2m (2) Frequency () (nu) : Frequency of a wave is defined as the number of waves which pass through a point in 1 second. It is measured in terms of Hertz (Hz ), second–1 , or cycle per second (cps) (1 Hertz = 1 second–1 ). 1 (3) Time period (T) : Time taken by a wave to pass through one point. T = second. (4) Velocity (c) : Velocity of a wave is defined as distance covered by a wave in 1 second. Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom c =   =  or = c/ or c = (s–1) ×  (m) or c =  (m s–1) Since c is constants i.e. frequency is inversely proportional to . (5) Wave number ( ) ( nu bar) : It is the reciprocal of the wave length i.e. number of waves present in 1cm. 1 =  It is measured in terms of cm–1, m–1 etc, (6) Amplitude (a) : The amplitude of a wave is defined as the height of crust or depth of trough. c  1 Important note : = = c  =        Chemistry Examples Example 3: The vividh Bharti station of All India Radio broadcast on a frequency of 1368 Kilo Hertz Calculate the wave length of the electromagnetic waves emitted by the transmitter. Solution : As we know velocity of light (c) = 3 × 108 m/s Given (frequency) = 1368 kHz = 1368 × 103 Hz = 1368 × 103 s–1 c 3  108ms−1    =  = = 219.3 m 1368  103 s−1 Example 4: Calculate in cm–1 and of yellow radiations have wavelength of 5800 Å. 8 1 1 10 Solution : As we know = = = cm–1 = 17241.44 cm–1  5800Å 5800 = c = 3 × 1010 cm s–1× 1.7×104 cm–1 = 3 × 1.7×1014 = 5.1×1014 s-1 Example 5: A Particular radio station broadcast at a frequency of 1120 Kilo Hertz another radio station broadcast at a frequency of 98.7 mega Hertz. What are the wave length of radiations from each station. c 3  108 ms−1 Solution : Station 1st  = = = 267.86 m 1120  103 s−1 c 3 108 ms−1 Station 2nd = = = 3.0395 m 98.7  106 s−1 Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom Example 6: How long would it take a radio wave of frequency 6 × 103 s–1 to travel from mass to the earth, a distance of 8 × 107 km ? Solution : Distance to be travelled from mass to earth = 8 × 107 km = 8 × 1010 m    Velocity of EM waves = 3 × 108 m/s Distance 8  1010 m  Time = = = 2.66 × 102 s = 4 min 26 s Velocity 3  108 m / s 2.5.2 Planck's Quantum Theory According to Planck's quantum theory : (1) The radiant energy emitted or absorbed by a body not continuously but discontinuously in the form of small discrete packets of energy and these packets are called quantum. (2) In case of light, the smallest packet of energy is called as 'photon' but in general case the smallest packet of energy called as quantum. Found. JEE/NEET : Class IX/X (3) The energy of each quantum is directly proportional to frequency of the radiation i.e. hc  = c  E  E = h or E=      h is proportionality constant or Planck's constant h = 6.626 × 10–37 kJ s or 6.626 × 10–34 J s or 6.626 × 10–27 erg s (4) Total amount of energy transmited from one body to another will be some integral multiple of energy of a quantum. E = nh Where n is an integer and n = number of quantum nhc E = nh = = nhc  Examples Example 7: Calculate the energy of a photon of sodium light of wave length 5.862 × 10 –16 m in Joules. Solution :  = 5.886 × 10–16 m, c = 3 × 108 m s–1 nhc E = nh or { n = 1 }  hc 1  6.6  10−34 J s  3  10 ms−1 6.6  3  E= = −16 =  10−10 J = 3.38 × 10–10 J  5.862  10 m 5.862 Example 8: Calculate the frequency & energy of a photon of wave length 4000 Å. Solution : (a) Calculation of frequency :  = 4000 Å = 4000 × 10–10 m Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom c 3  10 m / s 8  = = = 0.75 × 1015 s–1 = 7.5 × 1014 s–1  4  10−7 m (b) Calculation of energy : E = h = 6.626 × 10–34 joule second × 7.5 × 1014 s–1 = 4.96 × 10–19 joule Example 9: Calculate the  and frequency of a photon having an energy of 2 electron volt Solution :  1eV = 1.602 × 10–19 J  2eV = 3.204 × 10–19 J = E hc hc (a) Calculation of wavelength () : E = or  =  E 6.626  10−34 Js  3  108 ms −1 = −19 = 6.204 × 10–7 m 3.204  10 J −1 c 3  10 ms 8 (b) Calculation of frequency (): = = −7 = 0.48 × 1015 s–1= 4.8 × 1014 s–1  6.204  10 m   Chemistry Example 10: Which has a higher energy ? (a) A photon of violet light with wave length 4000 Å (b) A photon of red light with wave length 7000 Å −34 −1 hc 6.626  10 Js 3  10 ms 8 Solution : (a) Violet light :Eviolet = = = 4.97 × 10–19 joule  4000  10−10 m −34 −1 hc 6.626  10 Js 3  10 ms 8 (b) Red light : Ered = = = 2.8 × 10–19 joule  7000  10−10 m So, Eviolet > Ered Example 11: How many photons of lights having a wave length of 5000 Å are necessary to provide 1 joule of energy. −10 nhc E  1J  5000  10 m Solution :   E=  n= = = 2.5 × 1018 photons.  hc 6.626  10−34 J s  3  108 ms−1 CHECK YOUR LEARNING-4 Planck's Theory 1. A 1kw radio transmitter operates at a frequency of 800 Hz. How many photons per second does it emit? (1) 1.89  1021 (2) 1.89  1033 (3) 6.02  1023 (4) 2.85  1020 2. Select incorrect statements? (1) Every object emits radiations whose predominant frequency depends on its Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom temperature. (2) The quantum energy of a wave is proportional to its frequency. (3) Photons are quanta of light. (4) The value of the plancks constant depends on energy. 3. The wave number of a wave of light is 2.0 × 1014 cm–1. The frequency of this light is: (1) 6.6 × 103 s–1 (2) 6.6 × 10–3 s–1 (3) 6.0 × 1024 s–1 (4) 6.0 × 10–14 s–1 4. The energy of photon of reddish light having wavelength 660 nm is (h = 6.6 × 10–34 J s) (1) 1 × 10–19 J (2) 3.0 × 10–18 J (3) 1 × 1019 J (4) 3.0 × 10–19 J 5. The MRI (magnetic resonance imaging) body scanners used in hospitals operate with 400 MHz radio frequency energy. The wavelength corresponding to this radio frequency is? (1) 0.75 m (2) 0.75 cm (3) 1.5 m (4) 2 cm Found. JEE/NEET : Class IX/X 2.6 Bohr's Atomic Model Some Important formulae : Coulombic force =Centrifugal force =Angular momentum = mvr ⚫ It is a quantum mechanical model. This model was based on quantum theory of radiation and Classical law of physics. The important postulates on which Bohr's Model is based are the following : 1st Postulate : ⚫ Atom has a nucleus where all protons and neutrons are present. ⚫ The size of nucleus is very small and it is present at the centre of the atom. 2nd Postulate : ⚫ Negatively charged electron are revolving around the nucleus in the same way as the planets are revolving around the sun. ⚫ The path of electron is circular. ⚫ The attraction force (Coulombic or electrostatic force) between nucleus and electron is equal to the centrifugal force on electron. i.e. Attractive force towards nucleus = centrifugal force. 3rd Postulate : ⚫ Electrons can revolve only on those orbits whose angular momentum (mvr) is integral h multiple of. 2 h i.e. mvr = (n = Whole number) 2 Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom Where h = Planck's constant,  = Constant h h h h h ⚫ Angular momentum can have values such as ,2 ,3 ,4 ,5 ,........but can not 2 2 2 2 2 h h h have fractional values such as 1.5 , 1.2 , 0.5 ,................ 2 2 2 ⚫ The orbits in which electron can revolve are known as stationary Orbits. 4th Postulate : ⚫ By the expiry of time, if electons remains in the stationary state then it does not lose energy. Such a state is called as "GROUND STATE". 5th Postulate : ⚫ Each stationary orbit is associated with definite amount of energy therefore these orbits are also called as energy levels and are numbered as 1, 2, 3, 4, 5,....... or K, L, M, N, O,..... from the nucleus outwards.   Chemistry 6th Postulate : ⚫ The emission or absorption of energy in the form of radiation can only occur when electron jumps from one stationary state to another. ⚫ Energy is absorbed when electron Jumps from inner to outer orbit and is emitted when electron moves from outer to inner orbit. ⚫ When electron moves from inner to outer orbit by absorbing definite amount of energy the new state of the e– is said to be excited state of the electron. ⚫ If electrons is in 1st excited state it means now electron is exist in 2nd orbits. Similarly if electrons is in 2nd excited state it means now electron is exist in 3rd orbit and so on......... Ionization energy : The required energy to liberate an electron from the ground state of an isolated atom is called the ionization energy. Separation energy : Required energy to escape out an electron from its excited state is called as separation energy. 2.6.1 Application of Bohr's Model 2.6.1.1. Radius of Various Orbits (Shell) n2 rn = 0.529  Å Z This formula is only applicable for hydrogen and hydrogen like species i.e. species contains single electron. Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom Examples Example 12: Calculate the radius of 1st, 2nd, 3rd, 4th Bohr's Orbit of hydrogen. n2 Solution : Radius of Bohr's orbit r = 0.529 × Å Z 12 (a) Radius of Ist orbit : r = 0.529 × Å = 0.529Å 1 22 (b) Radius of IInd orbit : r = 0.529 × = 0.529 × 4 = 2.116Å 1 32 (c) Radius of IIIrd orbit : r = 0.529 × = 0.529 × 9 = 4.761Å 1 42 (d) Radius of IVth orbit : r = 0.529 × = 0.529 × 16 = 8.464Å 1 Found. JEE/NEET : Class IX/X Example 13: Calculate the radius ratio of 3rd & 5th orbit of He+. n2 Solution :  r = 0.529 × Å and Atomic Number of He = 2 Z (3) 2 9 (5)2  r3 = 0.529 × = 0.529 × and r5 = 0.529 × 2 2 2 25 = 0.529 × 2 ( 3) 2 r3 0.529  Therefore = 2 or r3 : r5 = 9 : 25 ( 5) 2 r5 0.529  2 Example 14: Calculate the radius ratio of 2nd orbit of hydrogen and 3rd orbit Li+2. Solution : Atomic number of H = 1, Atomic number of Li = 3, 22 2nd orbit radius of Hydrogen (r2)H = 0.529 × 1 32 3rd orbit radius of Li+2 ( r3 )Li +2 = 0.529 × 3 22 ( r2 )H 0.529   = 1 =4  ( r2 )H : ( r3 )Li =4:3 ( r3 )Li +2 32 3 +2 0.529  Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom 3 Example 15: The ratio of the radius of two Bohr's orbit of Li+2 is 1:9 What will be their nomenclature (1) K & L (2) L & M (3) K & M (4) K & N n2x 0.529  n 1 rx 1 3  nx 1 Solution : = = =  x = nx = 1 , n y = 3 ry 9 n2y ny 9 ny 3 0.529  3 Hence, orbit name is K and M Example 16: Calculate the radius of 2nd excited state of Li+2 Solution : 2nd excited state, means e– is present in 3rd shell so, 3 3 r3 = 0.529 × = 0.529 × 3Å = 1.587Å 3   Chemistry Example 17: Calculate the radius ratio of 2nd excited state of H & 1st excited state of Li+2 (3) 2 Solution : 2nd excited state, means e– is present in 3rd shell of hydrogen r3 = 0.529 × = 0.529 × 9 1 ( 2) 2 4 1st excited state, means e– exist in 2nd shell of Li+2 r2 = 0.529 × = 0.529 × 3 3 9 9 ( r3 )H 0.529  ( r3 )H 0.529  = 1 = = 1 = 27 ( r2 )Li +2 0.529  4 ( r2 )Li+2 0.529  4 4 3 3 2.6.1.2 Energy of an Electron 21.69×10−19 ×Z2 Z2 En = − J/atom or En = -13.6 × 2 eV /atom n2 n This formula is applicable for hydrogen atom & hydrogen like species i.e. single electron species. Shell O Since n can have only integral values, it follows that Shell N E5 total energy of the e– is quantised. Shell M E4 Shell L E3 The –ve sign indicats that the electron is under E2 Shell K attraction towards nucleus. E1 Energy difference between two energy levels : Nucleus + En2 − En1 = – 13.6 × Z2  12 − 12  Shell 1 n n  2 1  Shell 2 Shell 3 Energy level for H atom can be represented as follows : Shell 4 n = 6 or P E6 = – 0.38 eV Shell 5 Publishing\PNCF\2024-25\LIVE Module\SET-1\Foundation JEE-NEET\9th\Chemistry\2. Structure of Atom n = 5 or O E5 = –0.54 eV n = 4 or N E4 = – 0.85 eV E5 – E4 = 0.31 eV n = 3 or M E3 = – 1.51 eV E4 – E3 = 0.66 eV n = 2 or L E2 = – 3.4 eV E3 – E2