The modern quantum atom
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The Modern Quantum Atom. The nucleus and the discovery of the neutron What are electron-volts ? 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 Modern Quantum Atom

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The modern quantum atom

The Modern Quantum Atom

The nucleus and the discovery of the neutron

What are electron-volts ?

The Quantum atom


HW#8 posted (sorry)Prof. Artuso giving Monday’s class.I may allow a calc. For Exam 3

Rutherford s picture of the atom

Rutherford’s Picture of the Atom

Electrons circle the nucleus

due to the Coulomb force



~10-14 m

~10-11 m




This model was inspired by the results of scattering alpha-particlesoff of heavy nuclei (like gold, silver, etc). See previous lecture.

Rutherford Scattering:

James chadwick and the neutron

  • Performed a series of scattering experiments

  • with a-particles (recall a particles are He nucleus),

  • 42 He + 9 Be

12 C +10 n

James Chadwick and the Neutron

Circa 1925-1935

Picked up where Rutherford left off with more

scattering experiments… (higher energy though!)

  • Chadwick postulated that the emergent radiation

  • was from a new, neutral particle, the neutron.

  • Applying energy and momentum conservation

    he found that the mass of this new object was ~1.15 times that of the proton mass.


Awarded the Nobel Prize in 1935

Electron volts ev


-20 [kV]


1 kg


0 [V]


1 m

0 [J]


-20 [kV]

0 [kV]

***Electron-Volts (eV)***

  • When talking about subatomic particles, and individual photons,energies are very small (~10-12 or smaller).

  • It’s cumbersome to always deal with these powers of 10.

  • We introduce a new unit of energy, called the electron-volt (eV).

  • An [eV] is equivalent to the amount of energy a single electron gainswhen it is accelerated across a voltage of 1 [V].

  • Your TV tube accelerates electrons using 20,000 [V] = 20 [kV].

More on ev


More on [eV]

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] = 1.6x10-19 [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 3.2x10-15 [J] !

Even more on ev

Even more on [eV]


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] = 1.6x10-19 [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)”

Example 1

Example 1

A Cobalt-60 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] * (1.6x10-19 [J] / 1 [eV]) = 1.88x10-13 [J]

* Now, to get the wavelength, we use: E = hc/l, that is l = hc/E.

So, l = 6.63x10-34[J s]*3x108[m/s]/1.88x10-13 [J] = 1.1 x 10-12 [m]

* Now, convert [m] to [nm], 1.1 x 10-12 [m] * (109 [nm] / 1 [m]) = 1.1x10-3 [nm] It’s a GAMMA Ray

Example 2

Example 2

An electron has a mass of 9.1x10-31 [kg].

What is it’s rest mass energy in [J] and in [eV].

E = mc2 = 9.1x10-31*(3x108)2 = 8.2x10-14 [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]

Example 3

Example 3

A proton has a mass of 1.67x10-27 [kg].

What is it’s rest mass energy in [J] and in [eV].

E = mc2 = 1.67x10-27 *(3x108)2 = 1.5x10-10 [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]

Proton vs electron mass

Proton vs Electron Mass

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 = 9.1x10-31 [kg]

Proton mass = 1.67x10-27 [kg]Using [MeV/c2] as units of energy is easier…

Neils bohr and the quantum atom

Neils Bohr and the Quantum Atom

Circa 1910-1925

  • Pointed out serious problems with

  • Rutherford’s atom

  • Electrons should radiate as they orbit the

    nucleus, and in doing so, lose energy, until

    they spiral into the nucleus.

  • Atoms only emit quantized amounts of

    energy (i.e., as observed in Hydrogen spectra)

  • He postulated

  • Electric force keeps electrons in orbit

  • Only certain orbits are stable, and they do

    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.


Awarded the Nobel Prize in 1922

Bohr s picture of the atom




n =

Electronin lowest“allowed”energy level












Electronin excitedstate


Electron falls to the lowest energy level

Allowed Orbits

Bohr’s Picture of the Atom

Electrons circle the nucleus

due to the Electric force

Note: There are many more energy levels beyond n=5, they are omitted for simplicity

Atomic radiation


Electronin excitedstate(higher PE)


Electronin loweststate(lower PE)



n = 5

n = 5



n = 4

n = 4



n = 3

n = 3



n = 2

n = 2



n = 1

n = 1



Atomic Radiation

It is now “known” that when an electron is in an “excited state”,it spontaneously decays to a lower-energy stable state.

E5 > E4 > E3 > E2 > E1

  • The difference in energy, DE, is given by:DE = E5 – E1 = hn = Ephotonh = Planck’s constant = 6.6x10-34 [J s]

  • = frequency of light [hz] The energy of the light is DIRECTLY PROPORTIONAL to the frequency, n.Recall that the frequency, n, is related tothe wavelength by:c = n l (n = c / l)So, higher frequency  higher energy  lower wavelengthThis is why UV radiation browns your skinbut visible light does not !

One example could be:

Hydrogen atom energy levels






Hydrogen atom energy “levels”

Quantum physics provides the tools to compute the values ofE1, E2, E3, etc…The results are:

En = -13.6 / n2


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.

Example 4

Example 4

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 = 1.94x10-18 [J] / 6.6x10-34 [J s] = 2.9x1015 [1/s] = 2.9x1015 [hz]

l = c/n = (3x108 [m/s]) / (2.9x1015 [1/s]) = 1.02x10-7 [m] = 102 [nm]

This corresponds to low energy X-rays !

Some other quantum transitions

Some Other Quantum Transitions

This completed the picture or did it

This completed the picture, or did it…

  • Electrons were discovered ~1900 by J. J. Thomson

  • Protons being confined in a nucleus was put forth ~1905

  • Neutrons discovered 1932 by James Chadwick

  • Quantum theory of radiation had become“widely accepted”, although even Einstein had his doubts

  • Radiation is produced when atomicelectrons fall from a state of highenergy  low energy. Yields photonsin the visible/ X-ray region.

  • A nucleus can also be excited, andwhen it “de-excites” it also givesoff radiation  Typically gamma rays !

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