Novalgin Determination

Novalgin Determination

ELECTRONS IN ATOMS Dr Mehmet Gkhan ALAYAN 09/2019 Basic concepts A wave is a disturbance that transmits energy through space or a material medium. The distance between identical points on successive waves is called the wavelength

(lambda). lambda). The frequency v (lambda). nu) of the wave is the number of waves that pass through a particular point in one second. The amplitude is the vertical distance from the midline of a wave to the peak or trough. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves.

Speed of the wave speed = wavelength * frequency u=*v Wavelength is usually expressed in units of meters, centimeters, or nanometers, frequency is measured in hertz (lambda). Hz); 1 Hz = 1 cycle/s = 1 s-1

The electromagnetic spectrum The wavelength of electromagnetic radiation is shorter for high frequencies and longer for low frequencies. The visible region, extends from violet, 400nm, to red, 780 nm. Gamma rays have the shortest wavelength and highest frequency; radio waves have the longest wavelength and the lowest frequency

Question Most of the light from a sodium vapor lamp has a wavelength of 589 nm. What is the frequency of this radiation? Change the units of from nanometers to meters. v= c/ = 2.998* 108 m s-1 / 5.89 * 10-7 m = 5.09 * 1014 Hz Question

From the following colors of visible light, which is most energetic? (lambda). a) green, (lambda). b) red, (lambda). c) yellow. the higher the frequency, the more energetic the radiation. GREEN YELLOW RED Characteristic of Electromagnetic Waves

The Visible Spectrum The spectrum of white light (lambda). a) Dispersion of light through a prism. Red light is refracted the least and violet light the most when white light is passed through a glass prism. The other colors of the visible spectrum are found between red and violet. (lambda). b) Rainbow near a waterfall. Here, water droplets are the dispersion medium.

Atomic Spectra The visible spectrum is a continuous spectrum because the light being diffracted consists of many wavelength components. If the source of a spectrum produces light having only a relatively small number of wavelength components, then a discontinuous spectrum is observed. If the light source is an electric discharge passing through a gas, only certain colors are seen in the spectrum.

Or, if the light source is a gas flame into which an ionic compound has been introduced, the flame may acquire a distinctive color indicative of the metal ion present. These discontinuous spectra are called atomic, or line, spectra Plancks Quantum Theory Max Planck (lambda). 1858 1947) made a revolutionary proposal: Energy, like matter, is discontinuous.

Classical physics places no limitations on the amount of energy a system may possess, whereas quantum theory limits this energy to a discrete set of specific values. Planck gave the name quantum to the smallest quantity of energy that can be emitted (or absorbed) in the form of electromagnetic radiation. E = hv h = 6.62607 * 10-34 J s (lambda). Plancks constant) v= c/

E = h c / According to quantum theory, energy is always emitted in integral multiples of hv; for example, hv, 2 hv, 3 hv Question Calculate the energy (lambda). in joules) of (lambda). a) a photon with a wavelength of 3.75 * 104 nm (lambda). infrared region) and (lambda). b) a photon with a wavelength of 3.75 * 10-2 nm (lambda). X ray region).

a) E = h c / =(lambda). 6.63 * 10-34 J s)(lambda). 3.00 * 108 m/s) /(lambda). 3.75 * 104 nm) / (lambda). 1*10-9 m/1 nm) = 5.30* 10 -21 J b) 5.30* 10 -15 J Question For radiation of wavelength 242.4 nm, the longest wavelength that will bring about the photodissociation

of what is the energy of (a) one photon, and (b) a mole of photons of this light? a) v = c/ = 2.998 * 108 m s-1/ 242.4 * 10-9 m = 1.237 * 1015s-1 E = hv= 8.196 * 10-19 J/photon b) E = 8.196 * 10-19 J/photon * 6.022*1023 photons/mol Homework

1) The protective action of ozone in the atmosphere comes through ozones absorption of UV radiation in the 230 to 290 nm wavelength range. What is the energy, in kilojoules per mole, associated with radiation in this wavelength range? 2) Chlorophyll absorbs light at energies of 3.056 * 10 -19 J/ photon and 4.414 * 10-19 J/photon. To what color and frequency do these absorptions correspond?

Bohr Atom Acc. to the laws of classical physics, an e moving in an orbit of a hydrogen atom would experience an acceleration toward the nucleus by radiating away energy in the form of electromagnetic waves. Thus, such an e would quickly spiral into the nucleus and annihilate itself with the p. why this does not happen: Bohr postulated that the e is allowed to occupy

only certain orbits of specific energies. An e in any of the allowed orbits will not radiate energy, so will not spiral into the nucleus. Bohr showed that the energies that an e in hydrogen atom can occupy are given by where RH, the Rydberg constant for the hydrogen atom = 2.18*10-18 J. The number n is an integer called the principal quantum number; it has the values n 5 1, 2, 3, . . . .

Bohr model of H atom The nucleus is at the center, and the e in one of the discrete orbits. Excitation of the atom raises the electron to higher-numbered orbits, as shown with black arrows. Light is emitted when the electron falls to a lower-numbered orbit. Two transitions that produce lines in the Balmer series of the hydrogen spectrum

are shown in the approximate colors of the spectral lines. Transitions in the hydrogen spectrum The energy levels in the hydrogen atom. Each energy level

corresponds to the energy associated with an allowed energy state for an orbit. Energy-level diagram for the hydrogen atom Energy difference Ground state is the the ground lowest energy state of an atom

The stability of the electron diminishes for n = 2, 3, . . . Each of these levels is called an excited state, or excited level, which is higher in energy than the ground state. A hydrogen electron for which n is greater than 1 is said to be in an excited state. During emission, the electron drops to a lower energy state : nfinal (lambda). nf). The difference between the energies of the initial (lambda). ni)and final states is:

E = Ef - Ei Question Determine the wavelength of the line in the Balmer series of hydrogen corresponding to the transition from n=5 to n=2? E = 2.179 * 10-18 J(lambda). 1/52-1/22)= -4.576*10-19 J Ephoton= E = hv v= Ephoton/h = 4.576*10-19 Jphoton-1 /6.626*10-34 Jsphoton-1

=6.906 * 1014 s-1 =c/v =434.1 nm Homework What is the wavelength (lambda). in nanometers) of a photon emitted during a transition from ni= 6 to nf= 4 state in the H atom?

Wave properties De Broglies contribution: waves can behave like particles and particles can exhibit wavelike properties. =h/mu , m, and u are the wavelengths associated with a moving particle, its mass, and its velocity, respectively. Equation implies that a particle in motion can be treated as a wave, and a wave can exhibit the properties of a

particle. Question Calculate the wavelength of the particle in the following two cases: (lambda). a) Calculate the wavelength associated with a 5.50 x 10-2 kg tennis ball traveling at 45 m/s. (lambda). b) Calculate the wavelength associated with an electron (lambda). 9.1094 x 10-31 kg) moving at 45 m/s.

a) =h/mu = 2.7*10-34 m b) 1.6 * 10-5 m Ionization Energy of Hydrogen The Bohr model of the atom helps to clarify the mechanism of formation of cations. The model also works for hydrogen-like species, such

as the ions He+2, Li+2 which have only one electron. Question Determine the kinetic energy of the electron ionized from a Li+2 ion in its ground state, using a photon of frequency 5.000*1016 s-1. Ephoton = IE +Keelectron

Quantum Numbers and Electron Orbitals 3 quantum numbers are derived from the mathematical solution of the Schrdinger equation for the hydrogen atom. These numbers required to describe the distribution of electrons in hydrogen and other atoms. n : principal quantum number; determines the energy of an orbital; 1,2,3..

l :the angular momentum quantum number; determines the shape of the orbitals, 0, 1,2..n-1 ml :the magnetic quantum number; describes the orientation of the orbital in space -l, (lambda). -l +1).0(lambda). l-1),+l Quantum numbers A fourth quantum numberthe spin quantum number describes the behavior of a specific electron and

completes the description of e. ms= +1/2 or -1/2 Quantum Numbers and Atomic Orbitals the hydrogen 1s, 2s, and 3s orbitals surface diagrams of the three 2p

orbitals Boundary surface diagrams of the five 3d orbitals Question What is the total number of orbitals associated with the principal quantum number n = 3? For n=3, the possible values of l= 0, 1,2.

one 3s orbital (lambda). n=3, , l= 0, and ml= 0); three 3p orbitals (lambda). n=3, , l= 1, and ml= -1,0,1); five 3d orbitals (lambda). n=3, , l= 2, and ml= -2,-1,0,1,2). The total number of orbitals is 1 + 3 + 5 = 9. Question From the following sets of quantum numbers (lambda). n,l,ml,ms)identify the set that is correct, and state the orbital designation for those quantum numbers:

Question Write the four quantum numbers for an electron in a 3p orbital n, l, ml, ms Rules for Assigning Electrons to Atomic Orbitals

1. Each shell or principal level of quantum number n contains n subshells. Ex: n =2, then there are two subshells of angular momentum quantum numbers (lambda). l) 0 and 1. 2. Each subshell of quantum number l contains (lambda). 2l +1) orbitals. Ex: l=1 then there are three p orbitals. 3. No more than two electrons can be placed in each orbital. So, the max number of electrons is simply twice the number of

orbitals that are employed. 4. the maximum number of electrons that an atom can have in a principal level n can be found by the formula 2n 2. Question What is the maximum number of electrons that can be present in the principal level for which n=3? n =3, l= 0, 1, and 2

2*9 orbitals= 18 e-s. Check: 2n2 =18 Energies of Orbitals Orbital energy levels in the hydrogen atom.

Orbital energy levels in a manyelectron atom. Electron Configuration Electron config. For ground state e of hydrogen atom Orbital diagram

Multielectron atoms The shielding effect refers to the effect of inner-shell electrons in shielding or screening outer-shell electrons from the full effects of the nuclear charge. In effect the inner electrons partially

reduce the nuclear charge. Radial probability plots for the 1s, 2s, and 2p orbitals. The 1s electrons effectively shield both the 2s and 2p electrons from the nucleus. The 2s orbital is more penetrating than the 2p orbital.

The Pauli Exclusion Principle For many-electron atoms, the principle states electrons in an atom should have the same n,l, ml and ms values (lambda). that is, these two electrons are in the same atomic orbital), so they must have different values of ms. Only two electrons may occupy the same atomic orbital, and these electrons must have opposite spins.

Hunds Rule For carbon (lambda). Z = 6) is 1s22s22p2 which config. is true? exclusion principle Acc. to Paulies principle all of them are true. However; the answer is provided by Hunds rule; stating that the most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins.

C is the true answer. Summary of rules for electron configuration 1) Electrons occupy orbitals in a way that minimizes the energy of the atom. 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p 2) No two electrons in an atom can have all four quantum

numbers alike (lambda). the Pauli exclusion principle). 3) When orbitals of identical energy (lambda). degenerate orbitals) are available, electrons initially occupy these orbitals singly (lambda). Hunds rule). Keep in mind Diamagnetism and Paramagnetism

Paramagnetic substances are those that contain net unpaired spins and are attracted by a magnet. On the other hand, if the electron spins are paired, or antiparallel to each other, the magnetic effects cancel out. Diamagnetic substances do not contain net unpaired spins and are slightly repelled by a magnet. Li atom contains one unpaired e and the Li metal is therefore paramagnetic.

The Aufbau Process Aufbau process (lambda). German word: building up) is to assign electron configurations to the elements of periodic table in order of increasing atomic number. Electrons added to the shell of highest quantum number in the aufbau process are called valence electrons.

Electron config of some atoms Electron configurations and the periodic table Question Write (a) the electron configuration of mercury, and (b)

an orbital diagram for the electron configuration of tin. a) [Xe]4f145d106s2 b) Question Write the ground-state electron configurations for (lambda). a) sulfur (lambda). S) and (lambda). b) palladium (lambda). Pd), which is diamagnetic. a) Sulfur has 16 electrons. The noble gas core in this case is [Ne].

1s22s22p63s23p4 or [Ne]3s23p4. b) Palladium has 46 electrons. The noble-gas core in this case is [Kr]. [Kr] represents 1s22s22p63s23p64s23d104p6 Palladium: 1s22s22p63s23p64s23d104p64d10 Answer: [Kr]4d10

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