A-level Physics (Advancing Physics)/Risks, Doses and Dose Equivalents/Worked Solutions

'''1. A mobile phone emits electromagnetic radiation. 1.2 watts of power are absorbed per. kilogram. Assuming that the radiation is absorbed uniformly across a 5 kg head, what dose of radiation would be delivered to the head when making a 10-minute telephone call?'''

Total energy absorbed = 1.2 x 10 x 60 $$dose = \frac{\mbox{energy}}{mass} $$

Stick in the numbers and the dose is equal to 144 Gy. NB: This is wrong, because question says 1.2 Watts/kg already so you don't need to divide by 5kg.

Note to reader: There is currently something wrong with this question, since 144 Gy is enough to kill people. Mobile phones don't do that. I'm working on it. --Sjlegg (talk) 14:30, 6 April 2009 (UTC) Suggestion : A mobile phone does not emit ionizing radiation as the kinetic energy is to low. Therefore, the gray unit cannot be used for this exercise. The unit of measurement for absorption of electromagnetic energy by a body is the SAR.It measures the time rate of absorption of electromagnetic energy by a body. It is measured in W / kg.

The SAR can be determined from the electric field strength E in the body or from the rate of temperature rise (Δt).

2. What dose equivalent does this correspond to?

0 as EM radiatiom from mobile phones isn't ionising.

3. How many nuclei are there in 1 mg of Americium-241?

The mass of 1 Americium-241 nucleus is 241u, which corresponds to 241 x 1.66 x 10−27 = 400 x 10−27kg.

$$\frac{10^{-3}}{400 \times 10^{-27}} = 2.5 \times 10^{21}$$ NB: This calculation is just wrong- $$\frac{10^{-3}}{241}*Avogadro's number = 2.5 \times 10^{18}$$

'''4. An ham sandwich becomes contaminated with 1 &mu;g of Americium-241, and is eaten by an 80 kg person. The half-life of Americium-241 is 432 years. Given that Americium-241 gives off 5.638 MeV alpha particles, how long would it be before a dose equivalent of 6 Sv is absorbed, making death certain?'''

First calculate the activity:

$$A = \lambda N$$

$$t_{\frac{1}{2}} = \frac{\ln{2}}{\lambda}$$

$$\lambda = \frac{\ln{2}}{t_{\frac{1}{2}}}$$

$$A = \frac{N\ln{2}}{t_{\frac{1}{2}}} = \frac{2.5 \times 10^{21} \times \ln{2}}{432 \times 365.24 \times 24 \times 60 \times 60} = 1.27 \times 10^{11}\mbox{ Bq}$$ NB: The wrong N has been used here, questions says micro not milli.

Then calculate the power (energy per. second):

$$P = 1.27 \times 10^{11} \times 5.638 \times 10^6 \times 1.6 \times 10^{-19} = 0.115\mbox{ W}$$

Then calculate the dose per. second:

$$\frac{0.115}{80} = 1.43\mbox{ mGys}^{-1}$$

Multiply this by the quality factor for alpha particles (20) to get a dose equivalent per. second of 0.0287 Svs−1.

$$\frac{6}{0.0287} = 209\mbox{ s} = 3\mbox{ min }29\mbox{ s}$$

5. What assumptions have you made?

We assumed that the Americium was absorbed uniformly throughout the body, and that the activity remained constant. The latter is acceptable since this period of time is relatively short. The former is not acceptable.