Question

Tests on human subjects in Boston in 1965 and 1966, following the era of atomic bomb testing, revealed average quantities of about 2 pCi (7×10−2 dis/s) of plutonium radioactivity in the average person. Suppose each disintegration produces one alpha particle. If each alpha particle deposits 8×10−13J of energy and if the average person weighs 78 kg, calculate the number of rads of radiation in 5 yr from such a level of plutonium. 1 Bq=1 disintegration/s (dps) 1 Ci=3.7×1010 Bq 1 rad=0.01 Gy, 1 Gy=1 J/(kg tissue) 1 rem=0.01 Sv, 1 Sv=1 J/kg

Answer 1

Answer:

$\large \boxed{1 \times 10^{-5}\text{ rad}}$

Explanation:

1. Number of seconds in 5 yr

$\text{Time} = \text{5 yr} \times \dfrac{\text{365.25 da}}{\text{1 yr}} \times \dfrac{\text{24 h}}{\text{1 da}} \times \dfrac{\text{60 min}}{\text{1 h}}\times \dfrac{\text{60 s}}{\text{1 min}} = 1.58 \times 10^{8}\text{ s}$

2. Energy absorbed

$\text{Energy} = 1.58 \times 10^{8}\text{ s} \times \dfrac{7 \times 10^{-2}\text{ dis}}{\text{1 s}} \times \dfrac{8 \times 10^{-13}\text{ J}}{\text{1 dis}} = 8.8 \times 10^{-6} \text{ J}$

3. Total radiation received

$\text{Radiation} = \dfrac{8.8 \times 10^{-6} \text{ J}}{\text{78 kg}} \times \dfrac{\text{1 Gy}}{\text{1 J \cdot kg}^{-1}} \times \dfrac{\text{1 rad}}{\text{0.01 Gy}} = 1 \times 10^{-5} \text{ rad}\\\\\text{The average person receives \large \boxed{\mathbf{1 \times 10^{-5}}\textbf{ rad}} over five rears}$