mystery #3 emission spectra of elements · the line spectra of several elements. assuming atomic...
TRANSCRIPT
Tube filled with elemental gas.
Voltage can be applied across both ends,
this causes the gas to emit light
Mystery #3 Emission Spectra of Elements
Line Spectra
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7-
Figure 7.8
The line spectra of
several elements.
Assuming Atomic Theory is correct howdid scientists explain the emission phenomena?
How does the emission spectrum relatethe electronic structure of the H atom?
! (nm)for the visible series, n1 = 2 and n2 = 3, 4, 5, ...
The First Explanations Were Empirical
= RRydberg
equation-
1
!
1
n22
1
n12
! (nm) = 364.56!
n2
n2 ! 22
"
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!
In 1885 Balmer developed an empirical mathematical relationship between " and n for the visible lines of hydrogen
emission spectrum.
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:
Bohr’s Theory of the Hydrogen Atom
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
Bohr’s Model of the Hydrogen Atom (1913)
En = -RH( )
1
n2
n (principal quantum number) = 1,2,3,…
RH (Rydberg constant) = 2.18 x 10-18J
The radius of each orbit increases with n2
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
where i refers to the initial state and f to the final state.
Bohr’s Model of the Hydrogen Atom
i f
%E = RH( )1
n2
1
n2
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The Bohr explanation of the H spectral lines.
En = -RH( )
1
n2RH (Rydberg constant) = 2.18 x 10-18J
Ephoton = %Eatom
Ef = -RH ( )
1
n2f
Ei = -RH ( )
1
n2i
i f
%E = RH( )1
n2
1
n2
nf = 1
ni = 2
Ephoton = Ef - Ei
The Bohr explanation of the H spectral lines.
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Quantum staircase.
Ephoton = 2.18 x 10-18 J x (1/25 - 1/9)
Ephoton = %E = -1.55 x 10-19 J
! = 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J
! = 1280 nm
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.
Ephoton = h x c / !
! = h x c / Ephoton
i f
%E = RH ( )1
n2
1
n2Ephoton =
How and why is e- energy quantizedand 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.
While working on his PhD thesis, Louis DeBroglie
proposed that if energy behaved like a particle
then perhaps matter could behave like a wave.
! = wavelength (m)
v = velocity (m/s)
h = Planck’s constant
(6.626 & 10-34 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.
1924-- Wave-Particle Duality
Louis DeBroglie
! =h
mv
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?
! =h
m"
λ =h
mν=
6.626× 10!34 kg m2/s
(0.120 kg) (44.7 m/s)= 1.24× 10!34 m
Table 7.1 The de Broglie Wavelengths of Several Objects
Substance Mass (g) Speed (m/s) ! (m)
slow electron
fast electron
alpha particle
one-gram mass
baseball
Earth
9x10-28
9x10-28
6.6x10-24
1.0
120
6.0x1027
1.0
5.9x106
1.5x107
0.01
25.0
3.0x104
7x10-4
1x10-10
7x10-15
7x10-29
1.2x10-34
4x10-63
! = h /mv
SOLUTION:
Calculating the de Broglie Wavelength of an Electron
Find the deBroglie wavelength of an electron with a speed
of 1.00 x 106 m/s (electron mass = 9.11 x 10-31 kg; h =
6.626 x 10-34 kg*m2/s).
Knowing the mass and the speed of the electron allows to use the
equation ! = h/mv to find the wavelength.
! = 6.626x10-34kg*m2/s
9.11x10-31 kg x 1.00x106m/s
= 7.27 x 10-10 m
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.
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
The Heisenberg Uncertainty Principle
While working as a
postdoctoral
assistant with Niels
Bohr, Werner
Heisenberg
formulated the
uncertainty principle.
%x %p = h/4'%x = position uncertainty
%p = momentum uncertainty
(p = mv)
h = Planck’s constantWe can never precisely
know the location and the
momentum (or velocity or
energy) of an object.
This is only important for
very small objects.
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.
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”.