Atoms PDF

# 12. ATOMS ## Dalton's Model of Atom - Dalton postulated that matter is made of atoms. ## Thomson Model of Atom - J. J. Thomson proposed the structure for the atom, which was modified by Rutherford and later by Niels Bohr. - According to Thomson, atoms Consist of positively charged spheres in w...

# 12. ATOMS ## Dalton's Model of Atom - Dalton postulated that matter is made of atoms. ## Thomson Model of Atom - J. J. Thomson proposed the structure for the atom, which was modified by Rutherford and later by Niels Bohr. - According to Thomson, atoms Consist of positively charged spheres in which entire mass and positive charge of the atom is uniformly distributed. - Inside the sphere, the electrons are embedded like seeds in watermelon or like plums in a pudding. - The no. of es is equal to the no. of positive charges in the atom, thus the atom is electrically neutral. ## Limitations of Thomson Atom Model - It could not explain the spectral semes of hydrogen and other atoms. - It could not explain the large angle scattering of particles. ## Rutherford's Alpha-Ray Scattering Experiment - Rutherford's alpha-ray scattering experiment led to the discovery of the atomic nucleus. (Geiger and Marsden Experiment) ## Alpha Particle - An alpha particle is a helium nucleus with two protons and two neutrons. - An alpha particle has 4 units of mass and two units of positive charge. ## Rutherford's Atom Model ### Structure 1. Every atom consists of a tiny central core called the atomic nucleus, in which entire positive charge and almost entire mass of the atom are concentrated. 2. The size of the nucleus is of the order of $10^{-15}$ m, which is very small as compared to the size of the atom which is of the order of $10^{-10}$ m. 3. The atomic nucleus is surrounded by a certain number of electrons. Total -ve charge on the nucleus is equal to total positive charge on surrounding es, so an atom as a whole is electrically neutral. 4. The es are revolving around the nucleus in various circular orbits. The necessary centripetal force required by the e for revolution is provided by the electrostatic force of attraction between the es and the nucleus. ### Distance of Closest Approach - When an alpha particle approaches a nucleus, with its K.E (K), it experiences Coulombic repulsion and K.E reduces to zero. At a particular distance, the electric potential energy. This distance is known as distance of closest approach. - Charge on alpha particle, $q_1=+2e$ - Charge on nucleus, $q_2=+ze$ - Electric potential energy at distance $r_0$ $U= \frac{1}{4\pi \epsilon_0} \frac{(ze)(2e)}{r_0}$ - At distance of closest approach $r_0$ $KE=PE$ $\frac{1}{2} mv^2= \frac{1}{4\pi \epsilon_0} \frac{(ze)(2e)}{r_0}$ - The radius of nucleus must be smaller than $r_0$. ### Impact Parameter - The Impact parameter is defined as the perpendicular distance of the initial velocity vector of the aplha particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom. ### Limitations of Rutherford Atom Model 1. According to Rutherford, the es moving in the orbit has an accelerated motion, thus the es must radiate energy in the form of electromagnetic waves. As the e loses energy continuously, it must spiral inwards and finally fall into the nucleus. 2. As the es emit energy continuously, there should be a Continuous Spectrum, but we can observe only line Spectrum. ## Electron Orbits - The es are revolving. around the nucleus. The necessary centripetal force is provided by the electrostatic force of attraction blw e and nucleus. - $F_c=F_E$ - $\frac{mv^2}{r}= \frac{1}{4 \pi \epsilon_0} \frac{ze^2}{r^2}$ - For hydrogen atom ($z=1$) $\frac{mv^2}{r}= \frac{1}{4 \pi \epsilon_0} \frac{e^2}{r^2}$ $v = \sqrt{ \frac{1}{4 \pi \epsilon_0} \frac{e^2}{mr}}$ ## Bohr's Model of Hydrogen Atom ### Concepts - Bohr combined the classical ideas and quantum concepts to give the Bohr Model of hydrogen atom. ### Three Basic Postulates of Bohr Model 1. Every atom consists of a central core called nucleus, in which entire positive charge and almost entire mass of the atom are concentrated. A suitable number of es are revolving around the nucleus in circular orbits. The centripetal force required for revolution is provided by the electrostatic force of attraction blw es and the nucleus. 2. According to Bohr, e can revolve only in certain discrete non radiating orbits called stationary orbits, for which the total angular momentum of the revolving e is an integral multiple of $h/2\pi$ where h is plancks constant. Thus the angular momentum of the orbiting e is quantized. ie, for Stationary orbit. $mvr = nh/2\pi$ $n=1,2,3.....$ (principal quantum no). 3. The emission or absorption of energy occurs only when an e jumps from one of its stationary orbit to another. The difference in the total energy of e in the permitted orbits is absorbed when the e jumps from inner orbit to outer and emitted when e jumps from outer to the inner orbit. - If E1 is the total energy of e in the inner orbit and E2 is the total energy of e in the outer orbit. Then if the e jumps from outer to the inner orbit is $E_1 - E_2 = hv$. ### Radii of Bohr's Stationary Orbits. - According to the first postulate, $F=F_E$, ie, $mv^2/r = kze^2/r^2$. - From second postulate $mvr = nh/2\pi$ - $v = nh/2\pi mr$ - Substituting (2) in (1) $m(nh/2\pi mr)^2 = kz e^2/r^2$ $r = n^2 h^2 / 4 \pi^2 mkze^2$ - For hydrogen atom $Z=1$ $r = n^2 h^2 / 4 \pi^2 mke^2$ - ie, the stationary orbits are equally spaced. ### Velocity of e in Bohr's Stationary Orbit. - From first postulate, $mv^2/r = kze^2/r^2$ - $v = \sqrt{kze^2/mr}$ - From second postulate. $mvr = nh/2\pi$ - $v = nh/2\pi mr$ - Equating (1) and (2) $ \sqrt{kze^2/mr} = nh/2\pi mr$. - $v = 2\pi kze^2 / nh$. - For hydrogen atom z = 1, $V = 2 \pi ke^2 / nh$. - Orbital velocity of e in outer orbits is smaller as compared to its value in the inner orbits. ### Frequency of e in Bohr's Stationary Orbit. - It is the number of revolutions completed per second by the e- in a stationary orbit, around the nucleus. We have $V = r\omega = 2\pi r (f/T)$. - $f = V/2\pi r$. - $f = 2\pi kze^2 / nh. a1r$ ### Total Energy of Electron in Bohr's` Stationary Orbit. - $Total Energy (E) = K.E (K) + P.E (U)$ - $KE = 1/2 mv^2$ - From 1st postulate, $mv^2 = kze^2/r^2$ - $1/2 mv^2 = 1/2 kze^2/r$ = KE - PE, $U = $ potential x charge. - $U = -kze x - e/r$ - $PE = -kze^2/r$ - $T.E, E = K + U$ - $E = 1/2kze^2/r - kze^2/r$ - $E = - kze^2 / r$ - $ but, r = n^2 h^2 /4\pi^2 mkze^2$ - $T.E = -2\pi^2mk^2ze^4 / n^2h^2$ - $E = -13.6 Z^2 / n^2 ev$ - For hydrogen atom $Z=1$ - $E = -13.6 / n^2 ev$ - Total energy E is -ve, which means the e is bound to the nucleus. - When n = 1, E = -13.6ev is called ground state energy. - $ie, E_1 = -13.6ev$ ### Ionisation Energy - The minimum energy required to free the e from the ground state of hydrogen atom is 13.6 ev. - As n increases value of -ve energy decreases, ie, energy is progressively larger in the outer orbit. ### Origin of Spectral Lines - Rydberg Formula. - At room temperature, most of the atoms in hydrogen are in ground state. If sufficient amount of energy is given es are raised to the excited states. These es can fall back to lower energy by emitting photons of particular energy. - Let $E_n$ be the energy of e in nth orbit and $E_i$ in the ith orbit. - When e jumps from $n_2$ (outer) to $n_1$ (inner) orbit according to 3rd postulate. - $E_2 - E_1 = hv$ - $ie, hv = - 2\pi^2mk^2ze^4 / n_2^2h^2 + 2\pi^2mk^2ze^4 / n_1^2h^2$ - $\frac{h}{2} = \frac{2\pi^2mk^2ze^4}{h^2} [\frac{1}{n}^2 - \frac{1}{n}^2]$ - $λ = \frac{h^2}{2\pi^2mk^2e^4} [\frac{n_2^2 - n_1^2}{n_1^2n_2^2}]$ - 1/λ = wave number [number of complete waves in unit length] - $λ= \frac {h^2}{2\pi^2mk^2e^4} \frac{n_2^2 - n_1^2}{n_1^2n_2^2} = R_H \frac{n_2^2 - n_1^2}{n_1^2n_2^2}$ - $R_H$ = Rydberg Constant. - $ch^3 = V = \frac{1}{λ}$ - For hydrogen atom $Z=1$ - $V= \frac{1}{λ} = R [ \frac{1}{n_1^2} - \frac{1}{n_2^2}]$ - This is the Rydberg formula for hydrogen atom. - $R = 1.097 x 10^7/m^1$ ### Bohr's Explanation of spectral Series of Hydrogen atom. - When atomic gas of hydrogen at low pressure is excited by passing electric current, the emission spectrum is obtained. The emission spectrum consists of a few bright lines on darkbackground - Bohr explained this spectral series theoretically as follows. - **Lyman Series.** - When es jumps to the ground state $n_1 = 1$ from any excited states $n_2 =2, 3, 4, ...$. The Rydberg formula can be used, $V = R[ \frac{1}{1^2} - \frac{1}{n_2^2}] $, $n_2 = 2, 3, 4 .....$. - This spectral series line in the Ultra Violet Region. - **Balmer Series.** - When es jumps to second orbit $n_1 = 2$ (first excited state) from any outer orbits $n_2 = 3, 4, 5 .....$ . It can be written as $ V = R[ \frac{1}{2^2} - \frac{1}{n_2^2}]$, $n_2 = 3, 4, 5 .....$. - These spectral lines lie in the Visible region - **Paschen Series.** - When es jumps to the thrid orbit $n_1 = 3 $ from any outer orbits $n_2 = 4, 5, 6 .....$. It can be written as $V= R[ \frac{1}{3^2} - \frac{1}{n_2^2}]$, $n_2 = 4, 5, 6 .....$. - These spectral lines lie in the mfra red Region. - **Brackett Series.** - When es jumps to the 4th orbit $n_1 = 4$ from the outer orbits $n_2 = 5, 6, 7 .....$. It can be written as $V = R[ \frac{1}{4^2} - \frac{1}{n_2^2}]$, $n_2 = 5, 6, 7 .....$. - These Lines lie in the Infra red Region - **Pfund series.** - When es jumps to the 5th orbit $n_1 = 5 $ from the outer orbits $n_2 = 6, 7, 8 .....$. It can be written as $V = R[ \frac{1}{5^2} - \frac{1}{n_2^2}]$, $n_2 = 6, 7, 8 .....$. - These lines lie in the Infra red Region. ## ENERGY LEVEL DIAGRAM - A diagram which represents the total energies of electron in different stationary orbit of an atom is called energy level diagram. - The total energy of e in the nth orbit of hydrogen atom is $E = -2 \pi^2 mk^2 e^4 / n^2 h^2$ - On substituting Standard values, $E= -13.6 / n^2 ev$ - The energies of various stationary orbits are - $n=1$, $E_1 = -13.6 ev$ - $n=2$, $E_2 = -13.6 / 2^2ev = -3.4ev$ - $n=3$, $E_3 = -13.6 / 3^2 ev = -1.51ev$ - $n=4$, $E_4 = -13.6 / 4^2 ev = -0.85ev$ - $n=5$, $E_5 = -13.6 / 5^2 ev = -0.54 ev$ - $n=6$, $E_6 = -13.6 / 6^2ev = -0.37ev$ - $n=7$, $E_7 = -13.6 / 7^2ev = -0.28ev$ - Similarly if n = ∞ - $E_\infty = -13.6 / \infty^2 ev$ - $E_\infty = 0 ev$ ## Energy level diagram of hydrogen atom. - $n = ∞$, $0 ev$ - $n = 7$, $-0.28ev$ - $n = 6$, $-0.37ev$ - $n = 5$, $-0.54ev$ - $n = 4$, $-0.85ev$ - $n = 3$, $-1.5lev$ - $n = 2$, $-3.4ev$ - $n=1$, $-13.6ev$ ## de-Broglie's Explanation of Bohr's Second Postulate of Quantization. - According to Bohr's second postulate the angular momentum of the e orbiting around the nucleus is quantized ie, $mvr = nh/2\pi$, $n=1, 2, 3...$. - According to de Broglie, matter wave is associated with the es in the circular orbit. Hence the es in the stationary orbits contain an integral number of de Broglie waves. - For an e revolving in nth orbit of radius $r_n$, Total distance covered = $ 2\pi r_n $ (Circumference) Total length of the wave = $ nλ$. - For permissible orbit, $2\pi r_n = nλ$ - de Broglie wavelength, $λ = h/mv_n$, $V_n$=speed of e- in nth orbit. - ∴∴ $2\pi r_n = nh/mv_n$ - $mvr_n = nh/2\pi$ - ie, the angular momentum of the e revolving in nth orbit must be an integral multiple of $h/2\pi$, which is the quantum condition proposed by Bohr. ### Limitations of Bohis theory. 1. The Bohis theory is applicable only for simplest atom like hydrogen. 2. The theory does not explain why orbits of es are taken as Circular, while elliptical orbits are also possible. 3. Bohr's theory does not explain the fine Structure of spectral lines even in hydrogen atom. 4. The theory does not say anything about the relative intensities of spectral lines. 6. The wave properties of es are not considered. ## Excitation and Ionisation Potentials - By absorbing energy, the e can jump from an inner orbit to some outer orbit of higher energy. This is called excitation and atoms are said to be in excited state. - The minimum accelerating potential required for an e to jump from the ground state (n=1) to one of the outer orbits is called excitation potential OR Resonance potential. - Eg: If the e jump from ground state of hydrogen atom to first excited state (n=2). - $E_1 = -13.6ev$ - $E_2 = -3.4ev$ - Energy required for the jump, $E = -3.4-(-13.6)$ - The corresponding excitation potential = $10.2 V$ - Ionisation is the phenomenon of removal of an electron from the outermost orbit of an atom. - The minimum accelerating potential required to remove an e from the outermost orbit of the atom is called ionisation potential. - Eg: The total energy in ground state of hydrogen = -13.6ev. - To remove the e- from hydrogen atom energy required is +13.6 ev : Ionisation potential is +13.6V. - $V = 13.6 Z^2 / n_2^2 $ Volt.