http://phet.colorado.edu/en/simulation/quantum-tunneling
Application of Quantum
Tunneling:
Radioactive decay
George Gamow: 1928
Radioactive decay
(Quantum tunneling – George Gamow)
Nucleus is unstable → ejects alpha particle (2 netrons, 2 protons)
Typically found for large atoms with lots of protons and neutrons.
Polonium-210
84 protons,
126 neutrons
Proton (positive charge)
Neutron (no charge)
Radioactive decay
Proton (positive charge) In alpha-decay, an alpha-particle is
Neutron (no charge)
emitted from the nucleus.
Lead-206
82 protons,
124 neutrons
Polonium-210
84 protons,
126 neutrons
Nucleus has lots of protons and lots of
neutrons.
Two forces acting in nucleus:
- Coulomb force .. Protons really close
together, so very big repulsion from
coulomb force
- Nuclear force (attraction between nuclear
particles is very strong if very close
together) - called the STRONG Force.
How to figure out what's going on?
This raises the ratio of neutrons
to protons - makes for a more
stable atom.
(Neutrons are neutral.. no
coulomb repulsion, but nuclear
force attraction)
Potential energy curve for the α particle
Starting point: Always look at potential energy curve for particle!
KE
+
New nucleus
Nucleus
Alpha particle
(Z-2 protons,
(Z protons,
(2 protons,
& bunch of neutrons) bunch of neutrons) 2 neutrons)
Now look at this system- as the
distance between the alpha particle
and the nucleus changes.
KE
+
New nucleus
Nucleus
Alpha particle
(Z-2 protons,
(Z protons,
(2 protons,
& bunch of neutrons) bunch of neutrons) 2 neutrons)
Strong attractive force
(Nuclear forces)
V(r)
Look at this system- as the
distance between the alpha particle
and the nucleus changes.
As we bring the α particle closer to the core,
what happens to potential energy?
r
As we bring the α particle closer to the core,
what happens to potential energy?
Coulomb repulsion:
Nucleus:
(Z-2) protons
kq q
k ( Z − 2)(e)(2e)
V (r ) = 1 2 =
r
r
V=0 for r ∞
V (r ) =
kq1q2 k ( Z − 2)(e)(2e)
=
r
r
Wave function picture:
Energy
Energy
very small r (~1fm):
nuclear force dominates
~30 MeV
‘Large’ r: coulomb force dominates
V(r)
Exponential decay in the barrier
V(r)
1 to 10 MeV
~1-10MeV of KE
outside
Wave function of the free particle:
‘small’ KE Large wavelength
r
Edge of the nucleus (~8x10-15 m),
Nuclear (‘Strong’) force starts acting
strong attraction between nucleons.
Potential energy drops dramatically.
V(r)
Energy
Observations show Alpha-particles from the same
chemical element exit with a range of energies.
~100MeV
of KE inside
the nucleus
Wave function of the particle
inside the potential well: Large
KE small Wavelength
Solving Schrodinger
equation for this
potential energy is hard!
V(x)
Square barrier is much easierand get almost the same answer!
V(x)
9 MeV KE
4 MeV KE
Different KE in different isotopes
Isotope: Different types of atoms of the
same chemical element (same number of
protons but different numbers of neutrons.)
We have already seen isotopes of hydrogen
in this class: Hydrogen(1p, 0n), Deuterium
(1p, 1n) and Tritium (1p, 2n).
# neutrons influence nuclear potential
Nuclear Physics Sim
phet.colorado.edu/simulations/sims.php?sim=Alpha_Decay
Remember this picture?
Scanning tunneling microscope (STM)
Measure current
between tip and
sample
“See” |Ψ|2 of electrons!
The probability to
find an electron that
is trapped inside this
ring of atoms is
highest at the place,
where the square of
the amplitude of the
electron wave
function is largest.