Chapter 42: Q. 59 (page 1238)

The uranium isotope ${}^{238}U$ is naturally present at low levels in many soils. One of the nuclei in the decay series of ${}^{238}U$ is the radon isotope ${}^{222}R n$ , which decays by emitting a localid="1650487447445" $5.50 M e V$ alpha particle with localid="1650487458663" $t_{1 / 2} = 3.82$ days. Radon is a gas, and it tends to seep from soil into basements. The Environmental Protection Agency recommends that homeowners take steps to remove radon, by pumping in fresh air, if the radon activity exceeds $4 p C i p e r l i t e r$ of air.
a. How many ${}^{222}R n$ atoms are there in $1 m^{3}$ of air if the activity is $4 \frac{p C i}{L}$ ?
b. The range of alpha particles in air is $\approx 3 c m$ . Suppose we model a person as a $180 - c m$ -tall, $25 - c m$ -diameter cylinder with a mass of $65 k g$ . Only decays within $3 c m$ of the cylinder can cause exposure, and only $\approx 50 %$ , of the decays direct the alpha particle toward the person. Determine the dose in $m S v p e r y e a r$ for a person who spends the entire year in a room where the activity is $4 p C i / L$ .
c. Does the EPA recommendation seem appropriate? Why?

Short Answer

Expert verified

(a) No. of atoms are, $N = 7.05 \times 10^{7} a t o m s$

(b) Dose is, $3.3 \times 10^{- 2} m S v$

Step by step solution

Part (a) Step 1: Given information

We are given that,

Half life of alpha particle is, $t_{\frac{1}{2}} = 3.82 d a y s$

Activity is, $4 p C i / L$

Part (a) Step 2: Simplify

We know that,

Decay rate is given as,

$r = \frac{\ln (2)}{t_{\frac{1}{2}}} = \frac{\ln (2)}{330048} = 2.1 \times 10^{- 6} s^{- 1}$

Now, activity for $4 p C i / L$ in $1 m^{3}$

localid="1650487557849" $R = 4000 p C i = 4 \times 10^{- 9} C i \times \frac{3.7 \times 10^{10} B q}{1 C i} = 148 B q$

Therefore, No. of atoms are,

localid="1650487578188" $N = \frac{R}{r} = \frac{148 B q}{2.1 \times 10^{- 6} s^{- 1}} = 7.05 \times 10^{7} a t o m s$

Part (b) Step 1: Given information

We are given that,

Height of model of person is, $h = 180 c m$

Diameter of model of person is, $d = 25 c m$

Mass is, $m = 65 k g$

Part (b) Step 2: Simplify

As diameter is given so radius is,

$r_{0} = \frac{d}{2} = \frac{25}{2} = 12.5 c m = 0.125 c m$

Volume of air is,

$V = π {(R_{0})}^{2} (H) - π {(r_{0})}^{2} (h) = π {(0.155 m)}^{2} (1.83 m) - π {(0.125 m)}^{2} (1.80 m) = 0.0498 m^{3}$

Here, $R_{0}$ is radius of cylinder

And $H_{0}$ is height of cylinder

No. of decays per second is,

$R = r N = (2.1 \times 10^{- 6} s^{- 1}) (3.51 \times 10^{6}) = 7.4 \frac{d e c a y s}{s}$

No. of alpha particles is,

$N_{α} = \frac{1}{2} R \times (N o . o f s e c o n d s i n a y e a r) = \frac{1}{2} \times (7.4 d e c a y / s) \times 3.15 \times 10^{7} s / y r = 1.2 \times 10^{8} a l p h a / y r$

Dose is,

$d o s e = \frac{E}{m_{p}} \times \frac{1 G y}{1.00 J / k g} \times R B E = \frac{(1.2 \times 10^{8} a l p h a) (8.8 \times 10^{- 13} J / a l p h a)}{65 k g} \times \frac{1 G y}{1.00 J / k g} \times 20 = 3.3 \times 10^{- 2} m S v$

Part (c) Step 1: Given information

We need to find out that,

EPA recommendation is appropriate or not

Part (c) Step 2: Explanation

The dose is roughly $1 %$ of the natural background of radioactivity. This is due to EPA standards which is set to be $30 μ g / L$ .

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Short Answer

Step by step solution

Part (a) Step 1: Given information

Part (a) Step 2: Simplify

Part (b) Step 1: Given information

Part (b) Step 2: Simplify

Part (c) Step 1: Given information

Part (c) Step 2: Explanation

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