10.7 Medical applications and biological effects of nuclear radiation  (Page 6/18)

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Conceptual questions

Why is a PET scan more accurate than a SPECT scan?

Isotopes that emit $\alpha$ radiation are relatively safe outside the body and exceptionally hazardous inside. Explain why.

Alpha particles do not penetrate materials such as skin and clothes easily. (Recall that alpha radiation is barely able to pass through a thin sheet of paper.) However, when produce inside the body, neighboring cells are vulnerable.

Ionizing radiation can impair the ability of a cell to repair DNA. What are the three ways the cell can respond?

Problems

What is the dose in mSv for: (a) a 0.1-Gy X-ray? (b) 2.5 mGy of neutron exposure to the eye? (c) 1.5m Gy of $\alpha$ exposure?

Find the radiation dose in Gy for: (a) A 10-mSv fluoroscopic X-ray series. (b) 50 mSv of skin exposure by an $\alpha$ emitter. (c) 160 mSv of ${\beta }^{-}$ and $\gamma$ rays from the ${}^{40}\text{K}$ in your body.

$\text{Gy}=\frac{\text{Sv}}{\text{RBE}}$ : a. 0.01 Gy; b. 0.0025 Gy; c. 0.16 Gy

Find the mass of ${}^{239}\text{P}\text{u}$ that has an activity of $1.00\phantom{\rule{0.2em}{0ex}}\text{μCi}$ .

In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.

1.24 MeV

What is the dose in Sv in a cancer treatment that exposes the patient to 200 Gy of $\gamma$ rays?

One half the $\gamma$ rays from ${}^{99\text{m}}\text{T}\text{c}$ are absorbed by a 0.170-mm-thick lead shielding. Half of the $\gamma$ rays that pass through the first layer of lead are absorbed in a second layer of equal thickness. What thickness of lead will absorb all but one in 1000 of these $\gamma$ rays?

1.69 mm

How many Gy of exposure is needed to give a cancerous tumor a dose of 40 Sv if it is exposed to $\alpha$ activity?

A plumber at a nuclear power plant receives a whole-body dose of 30 mSv in 15 minutes while repairing a crucial valve. Find the radiation-induced yearly risk of death from cancer and the chance of genetic defect from this maximum allowable exposure.

For cancer: $\left(3\phantom{\rule{0.2em}{0ex}}\text{rem}\right)\left(\frac{10}{{10}^{6}\text{rem}·\text{y}}\right)=\frac{30}{{10}^{6}\text{y}},$ The risk each year of dying from induced cancer is 30 in a million. For genetic defect: $\left(3\phantom{\rule{0.2em}{0ex}}\text{rem}\right)\left(\frac{3.3}{{10}^{6}\text{rem}·\text{y}}\right)=\frac{9.9}{{10}^{6}\text{y}},$ The chance each year of an induced genetic defect is 10 in a million.

Calculate the dose in rem/y for the lungs of a weapons plant employee who inhales and retains an activity of $1.00\mu \text{Ci}$ ${}^{239}\text{Pu}$ in an accident. The mass of affected lung tissue is 2.00 kg and the plutonium decays by emission of a 5.23-MeV $\alpha$ particle. Assume a RBE value of 20.

The wiki-phony site states that the atomic mass of chlorine is 40 g/mol. Check this result. Hint: The two, most common stable isotopes of chlorine are: ${}_{17}^{35}\text{Cl}$ and ${}_{17}^{37}\text{Cl}$ . (The abundance of Cl-35 is $75.8\text{%}$ , and the abundance of Cl-37 is $24.2\text{%}$ .)

$\text{atomic mass}\phantom{\rule{0.2em}{0ex}}\left(\text{Cl}\right)=35.5\phantom{\rule{0.2em}{0ex}}\text{g/mol}$

A particle physicist discovers a neutral particle with a mass of 2.02733 u that he assumes is two neutrons bound together.

(a) Find the binding energy.

A nuclear physicist finds $1.0\phantom{\rule{0.2em}{0ex}}\text{μ}g$ of ${}^{236}\text{U}$ in a piece of uranium ore ( T 1/2 = $2.348\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\text{y}$ ). (a) Use the decay law to determine how much ${}^{236}\text{U}$ would had to have been on Earth when it formed $4.543\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\text{y}$ ago for $1.0\phantom{\rule{0.2em}{0ex}}\mu g$ to be left today. (b) What is unreasonable about this result? (c) How is this unreasonable result resolved?

a. $1.71\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{58}\phantom{\rule{0.2em}{0ex}}\text{kg}$ ; b. This mass is impossibly large; it is greater than the mass of the entire Milky Way galaxy. c. ${}^{236}\text{U}$ is not produced through natural processes operating over long times on Earth, but through artificial processes in a nuclear reactor.