# Types of radioactive decay

## Alpha

• He nucleus, 2 protons and 2 neutrons, charged +2
• Due to charge and mass, it is easy to stop
• One sheet of paper stops all alpha particles, even the most energetic
• Energy is quantized. A given isotope produce always alpha with the same energy
• Stopped by 1 inch of air
• Danger only when produced within the body, like if you ingested or breath.
• Quite difficult to detect with GM tubes, as they need an alpha-transparent window, like thin mica

## Beta-

• Are electrons, charged -1
• Much ligther than alpha, speed is relativistic
• Energy is distribution, since beta- is accompained by antineutrino. Energy is shared between the two.
• Blocked by few mm of aluminum.
• High energy beta interacts with high atomic weight and produce X rays. Better to stop using low atomic weight, like plastic (plexiglas), wood, water, … Bremsstrahlung radiation. Essentially charged particle is deflected by nuclei, lose some energy what is emitted as photon. Energy is similar to soft gamma, but spectrum is continuum.
• Easy to detect with GM tubes

Penetration r is the max depth the beta electrons can reach in material. After that, all electrons are adsorbed.

$r = \frac{0.412}{d} E^{1.29}$

d is density in g/cm3 (2.702 for aluminium)

E is the energy of the β rays in MeV

## Gamma

• It is electromagnetic wave, like visible light, UV or X-rays, microwave, radio, but more energetic
• Stopped by high atomic weight and high densities (-> so lead is good), but total mass per area in the radiation path is the most important.
• Pass through anything, adsorbed/shielded by high mass materials
• Detect with GM tubes easily
• Adsorpion follow exponential decay with thickness of material:

$N(x) = N_0 \exp(-\mu x)$

$\ln\left(\frac{N(x)}{N_0}\right) = -\mu x$

Half-Value Layer. Layer necessary to half the amount of incident gamma radiation

• Concrete: 44.5 mm
• Steel: 12.7 mm
• Lead: 4.8 mm
• Tungsten: 3.3 mm
• Uranium: 2.8 mm

Decrease quadratically with distance. Assuming point emission,

$I(r) = \frac{I_0}{4\pi r^2}$

so that

$\frac{I(r_1)}{I(r_2)} = \frac{r_2^2}{r_1^2}$

doubling the distance, the intensity decreases by 4.

Last update: 29 June 2018