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Transcript
```Quantum Physics and
Nuclear Physics
13.2 Nuclear Physics
Nuclear Physics
Determining the Mass of an Atom
If a charged particle moves at a constant speed
through a magnetic field it will experience a force
(=Bqv) that provides a centripetal force (=mv2/r)to
make it move through circular motion:
Thus
Bqv = mv2
r
m = Bqr
v
1
Clearly if velocity, charge and flux density are constant
then the mass of the particle is proportional to the radius
of the circle it moves through.
The Bainbridge Mass Spectrometer
In a Bainbridge mass spectrometer, positive ions
are created and injected into the velocity selector.
Here the positive ions are subjected to both
magnetic and electric fields. These create forces
that act in opposite directions.
Photographic
plate
If the two forces are balanced the particle will travel
in a straight line through the exit slit (s2 in the
diagram)...
Bqv = Eq
so the only particles to exit the velocity selector are
those with velocity equal to the ratio of the Electric
field strength (E) to the magnetic flux density (B)...
v= E
B
Sub into equation 1 gives:
m = B2qr
E
Q1.
The electric field strength between the plates of the
velocity selector in a Bainbridge mass spectrometer
is 1.20 x 105 Vm-1, and the magnetic field in both
regions is 0.600 T. A stream of singly charged neon
atoms moves in a circular path of 0.728 m radius in
the magnetic field. Determine
a. the mass of one neon atom
b. the mass number of this neon isotope.
(1 amu = 1.66 x 10-27kg)
a. 3.50 x 10-25 kg; b. 21
Q2.
In the Bainbridge mass spectrometer, suppose the
magnetic field B in the velocity selector is 1.30 T,
and ions having a speed of 4.00 x 106 ms-1 pass
through undeflected.
a) What is the electric field between the plates?
b) If the separation of the plates is 0.50 cm, what is
the potential difference across the plates?
c) If an ion of mass 5.30 x 10-26 kg passes into the
deflection chamber, calculate its radius of
deflection (assume q = 1.60 x 10-19C).
a) 5.20 x 106 V/m;
b) 2.60 x 104 V; c) 1.02m
Determining the Size of a Nucleus
When a positive charged particle approaches a
nucleus such as in scattering experiments (like
Rutherford’s alpha particle scattering experiment) it
is repelled by electrostatic forces.
The distance of closest approach (d) is taken as
being a measure of the radius of the nucleus (it has
the same order of magnitude).
α
v
α
d
Nucleus
Assuming all the initial KE of the alpha particle has
been converted into electric potential energy:
½ mv2 = kQq
r
so...
r = k Ze 2e
½ mv2
Q = charge on nucleus = Ze
q = charge on alpha = 2e
k = constant
( Z = proton number
e = charge on electron)
r = 4kZe2
mv2
As all particles position is uncertain, this gives only
a rough measure of the nuclear size.
Nuclear Energy Levels
It is not just the electrons that have energy in an atom.
The nucleus does too. When alpha particles are
emitted by a nucleus they are observed to have discreet
energies. Also, gamma photons are emitted with
discreet spectra.
These observations suggests that the nucleus must
have energy levels, similar to atomic energy levels.
We say a nucleus that is capable of emitting a gamma
(or alpha or beta) particle is unstable. This means it is
in an excited state, elevated to a higher nuclear energy
level. It can become more stable by emitting a gamma
particle and thus falling to an energy level closer to its
ground state.
This image shows how possible changes in the
excitation of a Nickel nucleus could lead to it
emitting gamma photons of discreet frequencies.
Note the difference
in the values of
energy compared to
atomic energy
levels.
Q. Explain the
differences in the
emission of X rays
and gamma rays
from the atom.
Beta Emission
Beta particles are not emitted with discreet
energies. They have a continuous range of energy
values:
To account for this in 1930 Wolfgang Pauli
postulated the existence of the neutrino ( - nu), not
detected until 1956.
When the nucleus falls to a lower energy level the
energy emitted is shared between the Beta particle
and the neutrino.
Thus the Beta
particle can
have a
continuous
range of
energies.
Beta plus decay
A beta plus particle is a positive electron (a positron,
a particle of anti-matter). When a proton turns into
a neutron in the nucleus the positron is emitted with
a neutrino ().
E.g.
22
11
Na →
22
10
Ne +
0
+1
e + 
Exponential Decay
Decay of a radioactive sample is an exponential
process. This means the rate at which the nuclei
decay (the activity (A) of the sample) is proportional
to the number of nuclei remaining at any instant.
Activity = -  N
dN
dt
=-N
N = N0 e– λt
We can also say...
 = decay constant (s-1)
nuclei at a certain time
nuclei at beginning (t=0)
A = A0 e– λt
Probability of Decay
The bigger the decay constant the quicker a
substance decays.
If N = 1 then
dN
dt
=-N
dN
dt
=-
- the decay constant gives us a measure of the
probability of any single nucleus decaying in unit
time.
Decay constant and half life
After t = t½ there will be half the original number of
radioactive nuclei. i.e. N = ½ N0
so ...
½ N0 = N0 e- λt½
½ = e- λt½
taking logs...
ln ½ = - λt½
this the same as...
ln 2 = λt½
t½ = ln 2
λ
Determination of Half Life
Experiment: Using a simulation to determine half life
Method:
- use a fresh sample of protactinium (a radioactive
nuclide with a short half life)
- determine values of activity every ten seconds.
- plot a suitable graph to determine the decay
constant and thus calculate half life.
Note: to plot a straight line graph take logs of both
sides of the activity equation:
A = A0 e– λt
ln A = ln A0 - λt
ln A = - λt + ln A0
( y = mx + c )
lnA
m=-λ
t
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