Bohr Model and Quantum Theory Lecture-2 Bohr Model and Quantum Theory

Bohr Atom The Planetary Model of the Atom Objectives: To describe the Bohr model of the atom. To explain the relationship between energy levels in an atom and lines in an emission spectrum.

Bohr’s Model Bohr’s Model Nucleus Electron Orbit Energy Levels

Photons Max Planck (1858-1947) Max Planck in 1900 stated that the light emitted by a hot object (black body radiation) is given off in discrete units or quanta. The higher the frequency of the light, the greater the energy per quantum.

Frequency (a) and (b) represent two waves that are traveling at the same speed. In (a) the wave has long wavelength and low frequency In (b) the wave has shorter wavelength and higher frequency

9-2. Photons The system shown here detects people with fevers on the basis of their infrared emissions, with red indicating skin temperatures above normal. In this way people with illnesses that may be infectious can be easily identified in public places.

Photons All the quanta associated with a particular frequency of light have the same energy. The equation is E = hf where E = energy, h = Planck's constant (6.63 x 10-34 J · s), and f = frequency. Electrons can have only certain discrete energies, not energies in between.

The Photoelectron Effect The photoelectric effect is the emission of electrons from a metal surface when light shines on it. The discovery of the photoelectric effect could not be explained by the electromagnetic theory of light. Albert Einstein developed the quantum theory of light in 1905.

What is light? Light exhibits either wave characteristics or particle (photon) characteristics, but never both at the same time. The wave theory of light and the quantum theory of light are both needed to explain the nature of light and therefore complement each other.

Bohr Model (1913) Assumptions 1) Only certain set of allowable circular orbits for an electron in an atom 2) An electron can only move from one orbit to another. It can not stop in between. So discrete quanta of energy involved in the transition in accord with Planck (E = h) 3) Allowable orbits have unique properties particularly that the angular momentum is quantized.

In circles the circumference (boundary of circle) is Energy of photon, E=hv         In case of particle, In circles the circumference (boundary of circle) is According wave model of atom,     Angular momentum,

Equations derived from Bohr’s Assumption Radius of the orbit Bohr Model (1913) Equations derived from Bohr’s Assumption Radius of the orbit r3 r2 h = Planck’s constant m = mass of electron e = charge on electron n = orbit number Z = atomic number r1 n=1 n=2 n=3

For He+ (also 1 electron) Bohr Model (1913) For H: For He+ (also 1 electron) Called Bohr radius Smaller value for the radius. This makes sense because of the larger charge in the center For H and any 1 electron system: n = 1 called ground state n = 2 called first excited state n = 3 called second excited state etc.

Bohr Model (1913) Which of the following has the smallest radius? First excited state of H Second excited state of He+ First excited state of Li+2 Ground state of Li+2 Second excited state of H

Bohr Model (1913) Which of the following has the smallest radius? First excited state of H Second excited state of He+ First excited state of Li+2 Ground state of Li+2 Second excited state of H

Problem Calculate the radius of 5th orbit of the hydrogen atom. n=5 h= 6.62 x 10-34 J sec m=9.109x10-31kg e=1.602x10-19C π=3.14 Z=1 r5= 13.225x10-10m

9-9. The Bohr Model Electron orbits are identified by a quantum number n, and each orbit corresponds to a specific energy level of the atom. An atom having the lowest possible energy is in its ground state; an atom that has absorbed energy is in an excited state.

Bohr Model (1913) Energy of the Electron constant, A = 2.18 x 10-18 J   constant, A = 2.18 x 10-18 J What’s happening to the energy of the orbit as the orbit number increases? Energy is becoming less negative, therefore it is increasing. The value approaches 0. Completely removed the electron from the atom.

Bohr Model (1913) + sign shows that energy was absorbed. xJ) = 1.64 x 10-18 J What is E when electron moves from n = 2 to n = 1? xJ) = - 1.64 x 10-18 J

So: Ephoton = |E|transition = h = h(c/) h = Planck’s constant = 6.62 x 10-34 J sec c = speed of light = 3.00 x 108 m/sec When E is positive, the photon is absorbed When E is negative, the photon is emitted

Problem A green line of wavelength 4.86x107 m is observed in the emission spectrum of hydrogen. a) Calculate the energy of one photon of this green light. b) Calculate the energy loses by the one mole of H atoms. Solution We know the wavelength of the light, and we calculate its frequency so that we can then calculate the energy of each photon. a) b)

What is the wavelength of the photon needed to move an electron from n = 1 to n = 2 in the H atom? recall: h = 6.62 x 10-34J sec,  = c 1.64 x 10-18 m 2.47 x 1015 m 6.62 x 10-34 m 1.21 x 10-7 m 2.18 x 10-18 m

What is the wavelength of the photon needed to move an electron from n = 1 to n = 2 in the H atom? So, light with this wavelength is absorbed When the electron goes back to n = 1, light with the same energy and wavelength will be emitted.

Ionization Energy energy needed to remove the outermost electron from an atom in its ground state. For H:electron moves from n = 1 to n = ∞ E∞= 0, E1 = - 2.18 x 10-18 J Therefore, the ionization energy for H is 2.18 x 10-18 J

DeBroglie Postulate (1924) Said if light can behave as matter, i.e. as a particle, then matter can behave as a wave. That is, it moves in wavelike motion. So, every moving mass has a wavelength () associated with it. where h = Planck’s constant v = velocity m = mass

What is the in nm associated with a ping pong ball (m = 2 What is the in nm associated with a ping pong ball (m = 2.5 g) traveling at 35.0 mph. 1.69 x 10-32 B) 1.7 x 10-32 C) 1.69 x 10-22 D) 1.7 x 10-23

Problem (a)Calculate the wavelength in meters of an electron traveling at 1.24 x107 m/s. The mass of an electron is 9.11x 10-28 g. (b) Calculate the wavelength of a baseball of mass 149g traveling at 92.5 mph. Recall that 1 J = 1 kgm2/s2.

b) m= 149g= 0.149kg

Heisenberg Uncertainty Principle To explain the problem of trying to locate a subatomic particle (electron) that behaves as a wave Anything that you do to locate the particle, changes the wave properties He said: It is impossible to know simultaneously both the momentum(p) and the position(x) of a particle with certainty

That’s all for today