Biological Effects of Radiation - 1st Year UG Physics

[P91]

Questions

Polonium-210 has a half life of 138.4 days. It decays to stable Pb-206 by emitting an alpha particle with a kinetic energy of 5.31 MeV. Alpha radiation of this energy has a relative biological effectiveness of 20.

a) If the difference in mass of Po-210 and Pb-206 nuclei is 4.00731u and the mass of an alpha particle is 4.00151u, estimate the fraction of the energy released in the radioactive decay of a Po-210 atom that is converted into kinetic energy of the alpha particle. What happens to the rest of the energy?

b) How many atoms are there in 1 microgram of Po-210?

c) Calculate the activity in becquerels of 1 microgram of Po-210

d) What is the total energy in joules of all of the alpha particles emitted by 1 microgram of Po-210 in one second?

e) If 1 microgram of Po-210 were ingested by a 70kg man, calculate the total equivalent radioactive dose in sieverts that he would receive in 48 hours. Comment on any assumptions that you make in your calculation.

f) Do you think that a sample of 1 gram of Po-210 isolated by Marie Curie in 1929 would still be a significant hazard today? Explain your reasoning

Relevant Equations

Dose equivalent in Sv = Absorbed dose in Gy x RBE
1u = 1.6605e-27 kg
Initial activity/initial number of atoms = time constant
half life = time constant * natural log(2)
half life = natural log (2) / decay rate
Number of atoms = Initial number of atoms * e^-(time of decay)*(decay rate)
dN/dt = - decay rate * number of atoms
Activity = decay rate * number of atoms

Attempt at a solution

a) I said that the initial energy of Po-210 is equal to the energy of Pb-206 + energy of alpha particle + kinetic energy. I rearranged to get Kinetic Energy(alpha) = change in mass * c^2 which gave my answer as 6e-10 Joules.

b) using the number of particles in a mole I divided the mass by the molar mass and multiplied by the number of particles. My answer was 2.86e15 atoms

c) using the equation below
half life = natural log (2) / decay rate
I plugged in my value of half life and found the decay rate and put that into the next equation.
Activity = decay rate * number of atoms
My answer was 1.66e9 Bq

d) Because the time period is one second, I divided the activity by the number of atoms in a mole of a substance. I then multiplied this by the molar mass of helium and my result was 1.11e-14 grams.

e) I have calculated the number of atoms emitted by using the initial amount and subtracting the amount at t = 48 hours. I now have a number for emitted particles in the man of 2.73e14 which has an energy of 1.638e5 Joules assuming that the particles have zero kinetic energy. I do not know how to convert this into the 'dose' that he would receive. I don't understand the relative biological effectiveness fully.

f) My initial thoughts are that I should calculate the number of atoms left from 1 gram, or the activity. If this is very small, which it must be since the half life is only 138.4 days - then it will not be a hazard today.

 

[misunderstood]

Hi,

I think youve misstyped the exponent for the activity..i got 1.66*10^8
Also part d) asks for the total energy of the emiited alpha particles, the energy of one alpha is given in the question as 5.31MeV. so it think this energy needs to multiplied by the number of alspha particles emitted in one second

 

[truth is life]

f) My initial thoughts are that I should calculate the number of atoms left from 1 gram, or the activity. If this is very small, which it must be since the half life is only 138.4 days - then it will not be a hazard today.

I don't think you need to do any calculations here. Just remember the definition of the half-life and the amount of time that's passed, then get the fraction of the original atoms that remain (you don't even need to do a precise calculation, just a Fermi estimate would be more than enough to show that it's harmless). The question says "explain your reasoning" not "do pointless calculations", after all.

 

[P91]

Hi,

I think youve misstyped the exponent for the activity..i got 1.66*10^8
Also part d) asks for the total energy of the emiited alpha particles, the energy of one alpha is given in the question as 5.31MeV. so it think this energy needs to multiplied by the number of alspha particles emitted in one second

I don't think you need to do any calculations here. Just remember the definition of the half-life and the amount of time that's passed, then get the fraction of the original atoms that remain (you don't even need to do a precise calculation, just a Fermi estimate would be more than enough to show that it's harmless). The question says "explain your reasoning" not "do pointless calculations", after all.

Thanks guys. I now have these as my answers:
a) 6e-10 J
b) 2.86e15 J
c) 1.66e8 Bq
d) 1.41e-4 J
e) 695 Sv
f) non-mathematical explanation

If anyone has worked them out, do your answers agree?

 

[paranormal]

However, if there is still a significant amount of activity, then it could still pose a hazard. This would also depend on the safety precautions taken with the sample and how it is stored. It is important to note that even small amounts of radioactive materials can be hazardous if not handled properly.