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Structure of Atoms Rutherford's model of the atom was a great advance, however, it does not give an satisfactory treatment of the electrons. To improve.

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Presentation on theme: "Structure of Atoms Rutherford's model of the atom was a great advance, however, it does not give an satisfactory treatment of the electrons. To improve."— Presentation transcript:

1 Structure of Atoms Rutherford's model of the atom was a great advance, however, it does not give an satisfactory treatment of the electrons. To improve the model insights from the study of light are Required. Combining the wave properties of light with that of particles it is possible to treat the behavior of the electron properly, using the new theory of Quantum Mechanics.

2 Electromagnetic Radiation Each wavelength of light can be use to investigate different properties of matter Spin Transition Rotational motion Electron Transitions Vibrational motion Nuclear Structure Scattering: Electronic Structure VRYBG

3 S A Waves are characterized by the properties: wavelength (λ): the distance between successive crests or successive troughs frequency (ν): the number of waves passing through a point in a given period of time amplitude (A): the height of a wave (from the node) speed (s) = wavelength × frequency The Wave 6 times 3 times node

4 Light Electric Field Magnetic Field Light is composed of an oscillating electric field perpendicular to an oscillating magnetic field.

5 Light S =  = C  = 2.998*10 8 m/s = Constant Plank’s Equation E = h  = h C / h  6.62618*10 -34 Js Exercise: Compute the energy of a radio wave with frequency at 500.00 MHz E = h  (6.62618*10 -34 Js)*(500.00*10 6 Hz) E = h  3.3131*10 -27 Js Hz = J s 1/s = J The energy of light is quantized as one h per wave packet

6 Particle Wave Duality Particle Wave Discrete energy Wave length Frequency Photon No Mass Momentum Diffraction

7 Exercise A laser emits light at a wavelength of 200 nm, for a period of 5.00 milliseconds onto a material that absorbs all the energy: i) Compute the energy of one photon? ii) How many photons were emitted? ii) How much energy was transferred to the material? i) Energy of one photon Need frequency  = C/ E = h E = h C/

8 E= (6.62618*10 -34 Js)(2.998*10 8 m/s) /(200*10 -9 m) E= 9.93 *10 -19 J ii) How many photons emitted in 5.00 ms?  = C/   2.998*10 8 m/s) /(200*10 -9 m) = 1.50*10 15 1/s (Hz)  # photons = t = ( 1.50*10 15 1/s)*(5.00*10 -3 s) = 7.5*10 12 Frequency is number of wave per second i.e number of photons per second. iii) Total energy? Total Energy = (# photons) * (Energy per photon) Total Energy = (7.5*10 12 )*(9.93 *10 -19 J)= 7.45 *10 -6 J


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