Radium 226 Disintegration

The Radium 226 nucleus

Assume that the activity of a $^{226}Ra$ (Radium 226 nucleus) source is $2.5 Ci$

a) At a distance of 50 cm away from this source, please find the flux of  energy (of the $\gamma$ emitted particles).  Consider that there is no attenuation and the radiation is isotropic.

b) Are the $\alpha$ (Helium nuclei) particles emitted  something to worry about? Can you say why, or why not?

Radium 226 decay
Radium Ra decays to Radon Rn directly or emitting a $\gamma$ photon

The radio-activity

From the figure $^{226}Ra$ (Radium 226) decays about the same as the figure shows for $^{225}Ra$, that is the energy of gamma quanta emitted is $E_0 =0.19 MeV$. The radio-activity of a source is measured in the International System in Bq (Bequerel). A derived unit is Curie (Ci). When we consider radiation on biological organisms we use the absorbed dose and the dose equivalent (measured in Sievert). The radio-activity is imprtant when we want to find the total energy emitted, by a source.

You can contiune reading on radio-activity here…

In conclusion an activity of

$1 Ci (Curie) =3.7*10^{10} Bq (Bequerel) = 3.7*10^{10} disintegrations/sec$

We know the energy of a $\gamma$ quanta so that the total energy in time unit is

$E = A*E0 =2.5*(0.19*10^6)*(1.6*10^{-19})*(3.7*10^{10}) =2.812*10^{-3} J$

When there is isotropic emission the radiation is contained inside a sphere. The surface of the sphere with radius R =50 cm (isotropic emission) is

$S =4*\pi*R^2 =3.14 m^2$

Hence we can find the energy flux as

$\Phi = E/S = (2.812*10^{-3})/3.14= 8.95*10^{-4} J/(m^*s) =8.95*10^{-4} W/m^2$

The $\alpha$ particles

The $\alpha$ particles are very easily stopped. As a result they are stopped even by a sheet of paper, or a few cm of air. This happens because they are very heavy, and even when they have high energy, their speed is low. Therefore at an average energy of let say 5 MeV, their speed is very small. This is why one do not has to worry about $\alpha$ emission.