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

 Page 5 / 18
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 ${}_{Z}^{A}\text{X}*\to {}_{Z}^{A}\text{X}+\gamma$

A round diaphragm S with diameter of d = 0.05 is used as light source in Michelson interferometer shown on the picture. The diaphragm is illuminated by parallel beam of monochromatic light with wavelength of λ = 0.6 μm. The distances are A B = 30, A C = 10 . The interference picture is in the form of concentric circles and is observed on the screen placed in the focal plane of the lens. Estimate the number of interference rings m observed near the main diffractive maximum.
A Pb wire wound in a tight solenoid of diameter of 4.0 mm is cooled to a temperature of 5.0 K. The wire is connected in series with a 50-Ωresistor and a variable source of emf. As the emf is increased, what value does it have when the superconductivity of the wire is destroyed?
how does colour appear in thin films
in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
derivation of lateral shieft
hi
Imran
total binding energy of ionic crystal at equilibrium is
How does, ray of light coming form focus, behaves in concave mirror after refraction?
Sushant
What is motion
Anything which changes itself with respect to time or surrounding
Sushant
good
Chemist
and what's time? is time everywhere same
Chemist
No
Sushant
how can u say that
Chemist
do u know about black hole
Chemist
Not so more
Sushant
DHEERAJ
Sushant
But ask anything changes itself with respect to time or surrounding A Not any harmful radiation
DHEERAJ
explain cavendish experiment to determine the value of gravitational concept.
Cavendish Experiment to Measure Gravitational Constant. ... This experiment used a torsion balance device to attract lead balls together, measuring the torque on a wire and equating it to the gravitational force between the balls. Then by a complex derivation, the value of G was determined.
Triio
For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?
as a free falling object increases speed what is happening to the acceleration
of course g is constant
Alwielland
acceleration also inc
Usman
which paper will be subjective and which one objective
jay
normal distributiin of errors report
Dennis
normal distribution of errors
Dennis
acceleration also increases
Jay
there are two correct answers depending on whether air resistance is considered. none of those answers have acceleration increasing.
Michael
Acceleration is the change in velocity over time, hence it's the derivative of the velocity with respect to time. So this case would depend on the velocity. More specifically the change in velocity in the system.
Big
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
give any fix value to wave length
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​.
Prema
what is physics?