Beta Spectroscopy


When a (parent) nucleus of mass number A = N + Z undergoes \beta decay, an electron (\beta- particle) is emitted and the daughter nucleus is characterized by the numbers A, Z+1 and N-1 (N is the number of neutrons, 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:

(1)^{A}_{Z}P_{N} \rightarrow ^{A}_{Z+1}D_{N-1} + e^- + \nu_e

For conservation of energy,

(2)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 m_P, m_D are the nuclear masses of the parent and daugther nucley, respectively and M_P, M_D are the atomic masses. Q is the so-called Q-value of the reaction and represents the total kinetic energy available for the daughter nucleus, the electron and the neutrino.

(3)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 Fig. 1.


Figure 1: Geometry of the \beta-spectrometer.

The magnetic field is produced by permament disk magnets of radius R. The electron path radius, r_e, equal to R/tan(\theta/2), is given by

(4)r_e = \frac{m_e |\vec{v}|}{e |\vec{B}|}

Therefore the momentum, p = |\vec{p}|, is given by

(5)p = m_e v = e B r_e

p = \frac{e B R}{\tan(\theta/2)}

where v = |\vec{v}| and B = |\vec{B}|. One obtains the kinetic energy (relativistically) from:

(6)K = \sqrt{ (pc)^2 + (m_e c^2)^2} - m_e c^2

where m_e c^2 = 0.511 MeV/c^2.


By altering the angle \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 ^{90}Sr source which decays as shoen in Fig. 2.


Figure 2: Decay scheme, masses and half lives of the ^{90}Sr \beta- -decay. The masses are in amu (931.49432  MeV/c^2)

Measure the radius, R, of the disk magnets, and use the gaussmeter to determine the average magnetic induction, B, between them. Note also the direction of \vec{B}. Place the Strontium (^{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 90^o positions.

Measure an entire angular distribution in steps of 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 5^o. Estimate the error of your background subtraced counts. Are 5 min enough ?


  1. For each angle, determine the number of counts per second above background (C/s) including their errors. (Negative values may be replaced by zero.)
  2. For each angle calculate the corresponding electron kinetic energy (see (6)).
  3. Using the expressions from (7), calculate for each angle \Delta K = \frac{dK}{d\theta}\cdot\Delta \theta with \Delta \theta = 10^o (make sure you use the right angle units!). Divide your count rate by the corresponding \Delta
K. As a result you have a count rate per unit kinetic energy.
  4. Plot a graph of the corrected count rate vs \beta particle kinetic energy (x-axis). Determine the 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 Q-values so obtained with those given by (3).

The geometry of the source and detector collimator of the spectrometer is such that it accepts a range of collection angles of about 10^o. This corresponds to a range of kinetic energies. This range can be calculated as follows using (5) and (6):

(7)\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)}