The Modern Quantum Atom. The nucleus and the discovery of the neutron What are electronvolts ? The Quantum atom. Announcements HW#8 posted (sorry) Prof. Artuso giving Monday’s class. I may allow a calc. For Exam 3. Rutherford’s Picture of the Atom. Electrons circle the nucleus
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The nucleus and the discovery of the neutron
What are electronvolts ?
The Quantum atom
Announcements
HW#8 posted (sorry)Prof. Artuso giving Monday’s class.I may allow a calc. For Exam 3
Electrons circle the nucleus
due to the Coulomb force
Corpuscles
(Electrons)
~1014 m
~1011 m
Positively
Charged
Nucleus
This model was inspired by the results of scattering alphaparticlesoff of heavy nuclei (like gold, silver, etc). See previous lecture.
Rutherford Scattering: http://galileoandeinstein.physics.virginia.edu/more_stuff/Applets/rutherford/rutherford.html
12 C +10 n
Circa 19251935
Picked up where Rutherford left off with more
scattering experiments… (higher energy though!)
he found that the mass of this new object was ~1.15 times that of the proton mass.
18911974
Awarded the Nobel Prize in 1935
ElectricPotential
20 [kV]
GPE
1 kg
10[J]
0 [V]
+
1 m
0 [J]

20 [kV]
0 [kV]
=1
How much energy does an electron gain when it is accelerated acrossa voltage of 20,000 [V] ?
E = 20,000 [eV]
[V] is a unit of “Potential”[eV] is a unit of Energy (can be converted to [J])
How can you convert [eV] to [J] ?
Not too hard… the conversion is: 1 [eV] = 1.6x1019 [J]
So, let’s do an example ! Convert 20 [keV] to [J].
Since the “k” == kilo = 1000 = 103, 20 [keV] = 20,000 [eV] = 2x104 [eV]
It’s a lot easier to say “20 [keV]” than 3.2x1015 [J] !
So, [eV] IS A UNIT OF ENERGY;
It’s not a “type” of energy (such as light, mass, heat, etc).When talking about energies of single photons, or of subatomic particles,we often use this unit of energy, or some variant of it.So,
1 [eV] = 1.6x1019 [J] (can be used to go back & forth between these two energy units)1 [keV] = 1000 [eV] = 103 [eV] “k = kilo (103)””
1 [MeV] = 1,000,000 [eV] = 106 [eV] “M = mega (106)”
1 [GeV] = 1,000,000,000 [eV] = 109 [eV] “G = giga (109)”
A Cobalt60 nucleus is unstable, and undergoes a decay where a 1173 [keV] photon is emitted. From what region of the electromagnetic spectrum does this come?
The energy is 1173 [keV], which is 1173 [keV] = 1173x103 [eV] = 1.173x106 [eV].
* First convert this energy to [J],
E = 1.173x106 [eV] * (1.6x1019 [J] / 1 [eV]) = 1.88x1013 [J]
* Now, to get the wavelength, we use: E = hc/l, that is l = hc/E.
So, l = 6.63x1034[J s]*3x108[m/s]/1.88x1013 [J] = 1.1 x 1012 [m]
* Now, convert [m] to [nm], 1.1 x 1012 [m] * (109 [nm] / 1 [m]) = 1.1x103 [nm] It’s a GAMMA Ray
An electron has a mass of 9.1x1031 [kg].
What is it’s rest mass energy in [J] and in [eV].
E = mc2 = 9.1x1031*(3x108)2 = 8.2x1014 [J]
Now convert to [eV]
What is an electron’s rest mass?
According to Einstein, m = E/c2, that is:
[mass] = [Energy] / c2
m = E / c2 = 0.51 [MeV/c2]
A proton has a mass of 1.67x1027 [kg].
What is it’s rest mass energy in [J] and in [eV].
E = mc2 = 1.67x1027 *(3x108)2 = 1.5x1010 [J]
Now convert to [eV]
What is a proton’s rest mass?
According to Einstein, m = E/c2, that is:
[mass] = [Energy] / c2
m = E / c2 = 940 [MeV/c2]
How much more massive is a proton than an electron ?
Ratio = proton mass / electron mass
= 940 (MeV/c2) / 0.51 (MeV/c2) = 1843 times more massive
You’d get exactly the same answer if you used: electron mass = 9.1x1031 [kg]
Proton mass = 1.67x1027 [kg]Using [MeV/c2] as units of energy is easier…
Circa 19101925
nucleus, and in doing so, lose energy, until
they spiral into the nucleus.
energy (i.e., as observed in Hydrogen spectra)
not radiate energy
Radiation is emitted when an e jumps from
an outer orbit to an inner orbit and the energydifference is given off as a radiation.
18851962
Awarded the Nobel Prize in 1922
Before
After
Radiatedphoton
n =
Electronin lowest“allowed”energy level
(n=1)
5
5
4
4
3
3
2
2
1
1
Electronin excitedstate
(n=5)
Electron falls to the lowest energy level
Allowed Orbits
Electrons circle the nucleus
due to the Electric force
Note: There are many more energy levels beyond n=5, they are omitted for simplicity
Energy
Electronin excitedstate(higher PE)
Energy
Electronin loweststate(lower PE)
E5
E5
n = 5
n = 5
E4
E4
n = 4
n = 4
E3
E3
n = 3
n = 3
E2
E2
n = 2
n = 2
E1
E1
n = 1
n = 1
Before
After
It is now “known” that when an electron is in an “excited state”,it spontaneously decays to a lowerenergy stable state.
E5 > E4 > E3 > E2 > E1
One example could be:
5
4
3
2
1
Quantum physics provides the tools to compute the values ofE1, E2, E3, etc…The results are:
En = 13.6 / n2
These results DO DEPEND ON THE TYPE OF ATOM OR MOLECULE
So, the difference in energy between the 3rd and 1st quantum state is:
Ediff = E3 – E1 = 1.51 – (13.6) = 12.09 (eV)
When this 3 1 atomic transition occurs, this energy is released in the form of electromagnetic energy.
In the preceding example, what is the frequency, wavelength of theemitted photon, and in what part of the EM spectrum is it in?
E = 12.1 [eV]. First convert this to [J].
Since E = hn n = E/h, so:
n = E/h = 1.94x1018 [J] / 6.6x1034 [J s] = 2.9x1015 [1/s] = 2.9x1015 [hz]
l = c/n = (3x108 [m/s]) / (2.9x1015 [1/s]) = 1.02x107 [m] = 102 [nm]
This corresponds to low energy Xrays !