10.7 Medical applications and biological effects of nuclear radiation  (Page 5/18)

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Typical doses received during diagnostic x-ray exams
Procedure Effective Dose (mSv)
Chest 0.02
Dental 0.01
Skull 0.07
Leg 0.02
Mammogram 0.40
Barium enema 7.0
Upper GI 3.0
CT abdomen 10.0

What mass of ${}^{137}\text{Cs}$ Escaped chernobyl?

The Chernobyl accident in Ukraine (formerly in the Soviet Union) exposed the surrounding population to a large amount of radiation through the decay of ${}^{137}\text{Cs}$ . The initial radioactivity level was approximately $A=6.0\phantom{\rule{0.2em}{0ex}}\text{MCi}.$ Calculate the total mass of ${}^{137}\text{Cs}$ involved in this accident.

Strategy

The total number of nuclei, N , can be determined from the known half-life and activity of ${}^{137}\text{Cs}$ (30.2 y). The mass can be calculated from N using the concept of a mole.

Solution

Solving the equation $A=\frac{0.693\phantom{\rule{0.2em}{0ex}}N}{{t}_{1\text{/}2}}$ for N gives

$N=\frac{A\phantom{\rule{0.2em}{0ex}}{t}_{1\text{/}2}}{0.693}.$

Entering the given values yields

$N=\frac{\left(6.0\phantom{\rule{0.2em}{0ex}}\text{MCi}\right)\left(30.2\phantom{\rule{0.2em}{0ex}}\text{y}\right)}{0.693}.$

To convert from curies to becquerels and years to seconds, we write

$N=\frac{\left(6.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{Ci}\right)\left(3.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{10}\phantom{\rule{0.2em}{0ex}}\text{Bq/Ci}\right)\left(30.2\phantom{\rule{0.2em}{0ex}}\text{y}\right)\left(3.16\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{7}\phantom{\rule{0.2em}{0ex}}\text{s/y}\right)}{0.693}=3.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{26}.$

One mole of a nuclide ${}^{A}\text{X}$ has a mass of A grams, so that one mole of ${}^{137}\text{Cs}$ has a mass of 137 g. A mole has $6.02\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{23}$ nuclei. Thus the mass of ${}^{137}\text{Cs}$ released was

$m=\left(\frac{137\phantom{\rule{0.2em}{0ex}}\text{g}}{6.02\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{23}}\right)\left(3.1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{26}\right)=70\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\text{g}=70\phantom{\rule{0.2em}{0ex}}\text{kg}.$

Significance

The mass of ${}^{137}\text{Cs}$ involved in the Chernobyl accident is a small material compared to the typical amount of fuel used in a nuclear reactor. However, approximately 250 people were admitted to local hospitals immediately after the accident, and diagnosed as suffering acute radiation syndrome. They received external radiation dosages between 1 and 16 Sv. Referring to biological effects in [link] , these dosages are extremely hazardous. The eventual death toll is estimated to be around 4000 people, primarily due to radiation-induced cancer.

Check Your Understanding Radiation propagates in all directions from its source, much as electromagnetic radiation from a light bulb. Is activity concept more analogous to power, intensity, or brightness?

power

Summary

• Nuclear technology is used in medicine to locate and study diseased tissue using special drugs called radiopharmaceuticals. Radioactive tags are used to identify cancer cells in the bones, brain tumors, and Alzheimer’s disease, and to monitor the function of body organs, such as blood flow, heart muscle activity, and iodine uptake in the thyroid gland.
• The biological effects of ionizing radiation are due to two effects it has on cells: interference with cell reproduction and destruction of cell function.
• Common sources of radiation include that emitted by Earth due to the isotopes of uranium, thorium, and potassium; natural radiation from cosmic rays, soils, and building materials, and artificial sources from medical and dental diagnostic tests.
• Biological effects of nuclear radiation are expressed by many different physical quantities and in many different units, including the rad or radiation dose unit.

Key equations

 Atomic mass number $A=Z+N$ Standard format for expressing an isotope ${}_{Z}^{A}\text{X}$ Nuclear radius, where r 0 is the radius of a single proton $r={r}_{0}{A}^{1\text{/}3}$ Mass defect $\text{Δ}m=Z{m}_{p}+\left(A-Z\right){m}_{n}-{m}_{\text{nuc}}$ Binding energy $E=\left(\text{Δ}m\right){c}^{2}$ Binding energy per nucleon $BEN=\frac{{E}_{b}}{A}$ Radioactive decay rate $-\frac{dN}{dt}=\lambda N$ Radioactive decay law $N={N}_{0}{e}^{\text{−}\lambda t}$ Decay constant $\lambda =\frac{0.693}{{T}_{1\text{/}2}}$ Lifetime of a substance $\stackrel{–}{T}=\frac{1}{\text{λ}}$ Activity of a radioactive substance $A={A}_{0}{e}^{\text{−}\lambda t}$ Activity of a radioactive substance (linear form) $\text{ln}\phantom{\rule{0.2em}{0ex}}A=\text{−}\lambda t+\text{ln}\phantom{\rule{0.2em}{0ex}}{A}_{0}$ Alpha decay ${}_{Z}^{A}\text{X}\to {}_{Z-2}^{A-4}\text{X}+{}_{2}^{4}\text{H}\text{e}$ Beta decay ${}_{Z}^{A}\text{X}\to {}_{Z+1}^{\phantom{\rule{1.5em}{0ex}}A}\text{X}+{}_{-1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}\stackrel{–}{v}$ Positron emission ${}_{Z}^{A}\text{X}\to {}_{Z-1}^{\phantom{\rule{1.5em}{0ex}}A}\text{X}+{}_{\text{+}1}^{\phantom{\rule{0.7em}{0ex}}0}\text{e}\phantom{\rule{0.2em}{0ex}}\text{+}\phantom{\rule{0.2em}{0ex}}v$ Gamma decay ${}_{Z}^{A}\text{X}*\to {}_{Z}^{A}\text{X}+\gamma$

as a free falling object increases speed what is happening to the acceleration
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Alwielland
acceleration also inc
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Rafi
40 cm into change mm
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
this msg is out of mistake. sorry friends​.
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