Mystery #3 Emission Spectra of Elements. Tube filled with elemental gas. Voltage can be applied across both ends, this causes the gas to emit light
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1 Mystery #3 Emission Spectra of Elements Tube filled with elemental gas. Voltage can be applied across both ends, this causes the gas to emit light
2 Line Spectra
3 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Figure 7.8 The line spectra of several elements. 7-
4 Assuming Atomic Theory is correct how did scientists explain the emission phenomena? How does the emission spectrum relate the electronic structure of the H atom?! (nm) for the visible series, n 1 = 2 and n 2 = 3, 4, 5,...
5 The First Explanations Were Empirical In 1885 Balmer developed an empirical mathematical relationship between " and n for the visible lines of hydrogen emission spectrum. λ (nm) = ( n 2 n ) In 1885 Johann Rydberg developed a generalized but empirical mathematical relationship between " and n for the all lines of hydrogen emission spectrum! This is freaky! Rydberg equation 1! = R 1 n n 2 2
6 Bohr s Theory of the Hydrogen Atom In 1913, scientist Neils Bohr provides a theoretical explanation of the emission spectrum of the hydrogen atom. An electron moves in a circular orbit about the proton. The electrostatic force of attraction between the nucleus and electron is balanced by the centrifugal force associated with the electron moving in the circular orbit about the nucleus. An electron could have only specific energies or orbits and they are quantized. This turns out to be the same as fixing the radii of the orbits. The electron can only have certain allowed energy states. Bohr showed the energies of these orbits are:
7 Bohr s Model of the Hydrogen Atom (1913) 1. e - can only have specific (quantized) energy values 2. light is emitted as e - moves from one energy level to a lower energy level 1 E n = -R H ( ) n 2 n (principal quantum number) = 1,2,3, R H (Rydberg constant) = 2.18 x J
8 Bohr s Model of the Hydrogen Atom The radius of each orbit increases with n 2 for n=1, we have the smallest Bohr Orbit for n=2, we have the next orbit which is 4x as big as the n=1 orbit Notice the negative energies calculated by the Bohr equation for n = 1, E = #Rhc for n = 2, E = # (1/4) Rhc for n = $, E = 0 Bohr postulated that light is emitted when an electron goes from an orbit with a high n value to a lower n value 1 1 n 2 %E = R H ( ) where i refers to the initial state and f to the final state. i n 2 f
9 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The Bohr explanation of the H spectral lines. 1 E n = -R H ( ) n 2 R H (Rydberg constant) = 2.18 x J
10 The Bohr explanation of the H spectral lines. E photon = %E atom n i = 2 E photon = E f - E i 1 E f = -R H ( ) n 2 f 1 E i = -R H ( ) n 2 i 1 1 n 2 %E = R H ( ) i n 2 f n f = 1
11 Quantum staircase. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
12 Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. E photon = E photon = 2.18 x J x (1/25-1/9) E photon = %E = x J E photon = h x c /! %E = R H ( )! = h x c / E photon! = 6.63 x (J s) x 3.00 x 10 8 (m/s)/1.55 x J! = 1280 nm 1 1 n 2 i n 2 f
13 How and why is e - energy quantized and what physics can explain it? Bohr s theory was doomed from the onset as classical physics laws mandate that electrons will collapse into the nucleus. While the work fit the data it had no good physical basis to stand on alone as a solid theory.
14 Wave-Particle Duality While working on his PhD thesis, Louis DeBroglie proposed that if energy behaved like a particle then perhaps matter could behave like a wave. Louis DeBroglie λ = h mv! = wavelength (m) v = velocity (m/s) h = Planck s constant (6.626 & J-s) This is only important for matter that has a very small mass and high velocity. In particular, the electron and small particles that move at high speeds.
15 de Broglie Wavelength What is the de Broglie wavelength (in meters) of a pitched baseball with a mass of 120. g and a speed of 100 mph or 44.7 m/s? λ = h mν λ = h mν = kg m 2 /s (0.120 kg) (44.7 m/s) = m
16 ! = h /mv Table 7.1 The de Broglie Wavelengths of Several Objects Substance Mass (g) Speed (m/s)! (m) slow electron 9x x10-4 fast electron 9x x10 6 1x10-10 alpha particle 6.6x x10 7 7x10-15 one-gram mass x10-29 baseball x10-34 Earth 6.0x x10 4 4x10-63
17 Calculating the de Broglie Wavelength of an Electron Find the debroglie wavelength of an electron with a speed of 1.00 x 10 6 m/s (electron mass = 9.11 x kg; h = x kg*m 2 /s). Knowing the mass and the speed of the electron allows to use the equation! = h/mv to find the wavelength. SOLUTION:! = 6.626x10-34 kg*m 2 /s = 7.27 x m 9.11x10-31 kg x 1.00x10 6 m/s
18 Wave Motion Can Be Restricted In Space Guitar strings (a) can only vibrate at certain wavelengths (frequencies). If we apply this situation to to electrons around a nucleus (b) then only certain whole number wavelengths would be allowed--- Quantization and other states would be disallowed.
19 Soon after de Broglie s theory it was found that e - (particles) could be diffracted (made to interfere like waves). x-ray diffraction of aluminum foil electron diffraction of aluminum foil
20 The Heisenberg Uncertainty Principle While working as a postdoctoral assistant with Niels Bohr, Werner Heisenberg formulated the uncertainty principle. We can never precisely know the location and the momentum (or velocity or energy) of an object. This is only important for very small objects. %x %p = h/4' %x = position uncertainty %p = momentum uncertainty (p = mv) h = Planck s constant The uncertainty principle means that we can never simultaneously know the position (radius) and momentum (energy) of an electron, as defined in the Bohr model of the atom.
21 The Uncertainty Principle!x!p " h 4# The act of measuring changes where the electron is. We can never know precisely where it is we can only give a probablity where to find it in space.
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