beta particle

Examples of beta particle in the following topics:

• Beta Decay

  • Beta decay is a type of radioactive decay in which a beta particle (an electron or a positron) is emitted from an atomic nucleus.
  • Beta decay is a type of radioactive decay in which a beta particle (an electron or a positron) is emitted from an atomic nucleus, as shown in .
  • Emitted beta particles have a continuous kinetic energy spectrum, ranging from 0 to the maximal available energy (Q), that depends on the parent and daughter nuclear states that participate in the decay.
  • The continuous energy spectra of beta particles occur because Q is shared between a beta particle and a neutrino.
  • Since the rest mass energy of the electron is 511 keV, the most energetic beta particles are ultrarelativistic, with speeds very close to the speed of light.

• Modes of Radioactive Decay

  • Alpha particles carry a positive charge, beta particles carry a negative charge, and gamma rays are neutral.
  • Alpha particles have greater mass than beta particles.
  • Their massive size (compared to beta particles, for instance) means alpha particles have very low penetration power.
  • Beta particles (β) have a higher penetration power than alpha particles (they are able to pass through thicker materials such as paper).
  • Beta particles can be stopped by aluminum shielding.

• Balancing Nuclear Equations

  • Common light particles are often abbreviated in this shorthand, typically p for proton, n for neutron, d for deuteron, α representing an alpha particle or helium-4, β for beta particle or electron, γ for gamma photon, etc.
  • This fits the description of an alpha particle.
  • This could also be written out as polonium-214, plus two alpha particles, plus two electrons, give what?
  • For the atomic number, we take 84 for polonium, add 4 (two times two) for helium, then subtract two (two times -1) for two electrons lost through beta emission, to give 86; this is the atomic number for radon (Rn).
  • Describes how to write the nuclear equations for alpha and beta decay.

• The Fields

  • We can use the potentials to determine the electric and magnetic fields produced by the moving particle.
  • $\displaystyle \beta \equiv \frac{\bf u}{c},~\text{so}~ \kappa = 1 - {\bf n} \cdot \beta$
  • It is important to remember that all of the properties of the particle are evaluated at the retarded time.
  • A few things to notice are that if the particle is not accelerating the electric field points to the current not the retarded position of the particle.
  • This allows us to graphically depict the field for a particle that is stopped suddenly.

• B.4 Chapter 4

  • The particle bounces off of the mirror.
  • The particle bounces off of the mirror.
  • $\displaystyle p^\mu = \left [ \begin{array}{c} m c \frac{1+\beta^2}{1-\beta^2} \\ -mc \frac{2\beta}{1-\beta^2} \\ 0 \\ 0 \end{array} \right ]$
  • Compare the energy of the particle in step (4) to the energy of the particle in step (9).
  • Has the energy of the particle increased?

• Motion in a magnetic field

  • Syhchrotron radiation, a.k.a. magnetic bremmstrahlung, is produced by relativistic charged particles travelling through a magnetic field.
  • and the particle gyrates around the magnetic field with a frequency
  • The acceleration ($\omega_B v_\perp$) is perpendicular to the motion of the particle so we can use formula (44) from Unit 3 to get the total power,
  • Let's assume that the particles have a random distribution of velocities relative to the direction of the magnetic field, so we need the mean value of
  • $\displaystyle \left \langle \beta^2_\perp \right \rangle = \frac{\beta^2}{4\pi} \int \sin^2\alpha d \Omega = \frac{2\beta^2}{3}$

• Transformation of Radiative Transfer

  • $\displaystyle p'_t = \gamma \left ( p_t - \beta p_x \right ) \\ p'_x = \gamma \left ( p_x - \beta p_t \right ) \\ p'_y = p_y \\ p'_z = p_z .$
  • $\displaystyle \gamma \left ( 1 - \beta \frac{p_x}{p_t} \right ) = \frac{\gamma \left ( p_t - \beta p_x \right )}{p_t} = \frac{p_t'}{p_t}.$
  • $\displaystyle \gamma \left (x_A - \beta t_A \right ) = v' \gamma \left (t_A - \beta x_A\right ) + x_A(0)$
  • $\displaystyle x_A = \frac{\beta + v'}{1+\beta v'} t_A + \frac{x_A'(0)}{\gamma \left (1 + \beta v'\right )}.$
  • Notice how the particle travels at a different velocity in the new frame and the relativistic addition of velocities.Now we find that

• Cherenkov Radiation

  • $\displaystyle \frac{d W}{d\omega d\Omega} = \frac{q^2 \epsilon^{1/2} \beta^2 \sin^2 \theta}{c} \left | \delta ( 1 - \epsilon^{1/2} \beta \cos \theta)\right |^2$
  • where $\theta$ is measured relative to the velocity of the particle.
  • The total energy radiated diverges; this simply results from our assumption that the charge travels through the dielectric material forever and this assumption is easy to relax by replacing the infinite integral with one over a time $ 2T$ during which the particle travels through the dielectric
  • $\displaystyle \frac{d W}{d\omega} = \frac{q^2 \omega}{c^2} \sin^2 \theta_c \left (2 c \beta T \right ) = \frac{q^2 \omega}{c^2} \left [ 1 - \frac{1}{\beta^2 \epsilon(\omega)} \right ] \left ( 2 c \beta T \right )$
  • where $2 c \beta T$ is the thickness of the dielectric region.

• Problems

  • The particle bounces off of the mirror.
  • Compare the energy of the particle in step (d) to the energy of the particle in step (9).
  • $\beta \approx 1 - \frac{1}{2\gamma^2}$By what factor does the energy of the particle increase each time it goes back and forth.
  • Write out the Lorentz transformation matrix for a boost in the x−direction to velocity $\beta_x$ followed by boost to a velocity $\beta_y$ in the y−direction.
  • Write out the Lorentz transformation matrix for a boost in the x−direction to velocity $\beta_x$, followed by boost to a velocity $\beta_y$ in the y−direction, followed a boost in the x−direction to velocity −$\beta_x$ followed by boost to a velocity −$\beta_y$ in the y−direction.

• How about relativistic particles?

  • $\displaystyle {\bf S} = {\bf n} \frac{q^2}{4\pi c \kappa^6 R^2} \left | {\bf n} \times \left \{ \left ( {\bf n} - \beta\right ) \times {\dot{\beta}} \right \} \right |^2$
  • $\displaystyle \frac{dP(t')}{d\Omega} = \frac{q^2}{4\pi c} \frac{ \left | {\bf n} \times \left \{ \left ( {\bf n} - \beta \right ) \times {\dot{\beta}} \right \} \right |^2}{\left ( 1 - {\bf n} \cdot \beta \right )^5} $
  • Let's start by assuming the $\beta$ is parallel to ${\dot{\beta}}$, so $\beta \times {\dot{\beta}}=0$.
  • Let's repeat the calculation for circular motion, in which $\beta \perp {\dot{\beta}}$.
  • If the particles are ultra-relativistic it is appropriate to neglect the parallel contribution completely.