Beta Spectroscopy
=================
Background
----------
When a (parent) nucleus of mass number :math:`A = N + Z` undergoes
:math:`\beta` decay, an electron (:math:`\beta`\ - particle) is
emitted and the daughter nucleus is characterized by the numbers
:math:`A`, :math:`Z+1` and :math:`N-1` (:math:`N` is the number of
neutrons, :math:`Z`, the atomic number, is the number of protons,).
Because the mass of a nucleus greatly exceeds that of the electron,
the daughter acquires virtually no kinetic energy. We might therefore
expect the electrons to be monoenergetic, with an energy equal to the
difference in rest energies of the parent and daughter nuclei. In
this experiment you will discover that the electron energies cover a
range of values, indicating the involvement of another particle, the
elusive antineutrino, which carries off some of the energy. The
complete process is represented by:
.. math::
:label: r1
^{A}_{Z}P_{N} \rightarrow ^{A}_{Z+1}D_{N-1} + e^- + \nu_e
For conservation of energy,
.. math::
:label: e1
m_P c^2 = m_D c^2 + m_e c^2 + Q
(M_P - Z M_e) c^2 = (M_D - (Z+1)m_e) c^2 + m_e c^2 + Q
where :math:`m_P, m_D` are the nuclear masses of the parent and
daugther nucley, respectively and :math:`M_P, M_D` are the atomic
masses. :math:`Q` is the so-called :math:`Q-`\ value of the reaction
and represents the total kinetic energy available for the daughter
nucleus, the electron and the neutrino.
.. math::
:label: e2
Q = (M_P - M_D) c^2
By measuring the radius of curvature of the electron path in a
magnetic field, the electron energy may be determined. The geometry
of the beta spectrometer you will use is shown in
:ref:`Fig. 1.`
.. _bs_fig1:
.. figure:: pictures/beta_spectroscopy.png
:align: center
:scale: 40 %
Figure 1: Geometry of the :math:`\beta`\ -spectrometer.
The magnetic field is produced by permament disk magnets of radius :math:`R`. The
electron path radius, :math:`r_e`, equal to :math:`R/tan(\theta/2)`,
is given by
.. math::
:label: e3
r_e = \frac{m_e |\vec{v}|}{e |\vec{B}|}
Therefore the momentum, :math:`p = |\vec{p}|`, is given by
.. math::
:label: e4
p = m_e v = e B r_e
p = \frac{e B R}{\tan(\theta/2)}
where :math:`v = |\vec{v}|` and :math:`B = |\vec{B}|`. One obtains the
kinetic energy (relativistically) from:
.. math::
:label: e5
K = \sqrt{ (pc)^2 + (m_e c^2)^2} - m_e c^2
where :math:`m_e c^2 = 0.511 MeV/c^2`.
Experiment
----------
By altering the angle :math:`\theta`, the electron detector (GM tube)
will record the number of electrons of different kinetic energies
incident on it. The maximum kinetic energy observed will correspond
to the situation where the kinetic energy of the antineutrino is zero.
This maximum kinetic energy is then equal to the Q value of the
process.
This experiment uses a :math:`^{90}Sr` source which decays as shoen in
:ref:`Fig. 2`.
.. _bs_fig2:
.. figure:: pictures/Sr90_decay.png
:align: center
:scale: 40 %
Figure 2: Decay scheme, masses and half lives of the
:math:`^{90}Sr` :math:`\beta`- -decay. The masses are in amu
(:math:`931.49432 MeV/c^2`)
Measure the radius, :math:`R`, of the disk magnets, and use the gaussmeter to
determine the average magnetic induction, :math:`B`, between them. Note also
the direction of :math:`\vec{B}`. Place the Strontium (:math:`^{90}Sr`) source under the
spring clip in the source arm. Make sure that the center of the
source is centered on the hole in the aluminum. Set the source arm at
one of the :math:`90^o` positions.
Measure an entire angular distribution in steps of :math:`10^o`. Count
at each point at least for 5 min. Make a background measurement for at
least 10 min. If the time permits it, measure additional points to have
an angular distribution with a step size of :math:`5^o`. Estimate the
error of your background subtraced counts. Are 5 min enough ?
Analysis
--------
#. For each angle, determine the number of counts per second above
background (C/s) including their errors. (Negative values may
be replaced by zero.)
#. For each angle calculate the corresponding electron
kinetic energy (see :eq:`e5`).
#. Using the expressions from :eq:`e6`, calculate for each angle
:math:`\Delta K = \frac{dK}{d\theta}\cdot\Delta \theta` with
:math:`\Delta \theta = 10^o` *(make sure you use the right angle
units!)*. Divide your count rate by the corresponding :math:`\Delta
K`. As a result you have a count rate per unit kinetic energy.
#. Plot a graph of the corrected count rate vs :math:`\beta` particle
kinetic energy (x-axis). Determine the :math:`Q`\ -value of the
reaction. Because of the rather crude equipment used, your graphs
will probably show a lot of tailing at the higher energies. Ignore
this and obtain an intercept with the x-axis using the relatively
linear portion of the graph. Compare your :math:`Q`\ -values so obtained
with those given by :eq:`e2`.
The geometry of the source and detector collimator of the spectrometer
is such that it accepts a range of collection angles of about
:math:`10^o`. This corresponds to a range of kinetic energies. This
range can be calculated as follows using :eq:`e4` and :eq:`e5`:
.. math::
:label: e6
\frac{dK}{d\theta}= \frac{dK}{dp}\cdot\frac{dp}{d\theta}
\frac{dK}{dp} = \frac{pc^2}{\sqrt{p^2c^2 + m_e^2c^4}}
\frac{dp}{d\theta} = \frac{-eBR}{2\sin^2(\theta/2)}