<< Chapter < Page Chapter >> Page >

Human-made (or artificial) radioactivity has been produced for decades and has many uses. Some of these include medical therapy for cancer, medical imaging and diagnostics, and food preservation by irradiation. Many applications as well as the biological effects of radiation are explored in Medical Applications of Nuclear Physics , but it is clear that radiation is hazardous. A number of tragic examples of this exist, one of the most disastrous being the meltdown and fire at the Chernobyl reactor complex in the Ukraine (see [link] ). Several radioactive isotopes were released in huge quantities, contaminating many thousands of square kilometers and directly affecting hundreds of thousands of people. The most significant releases were of 131 I , 90 Sr , 137 Cs , 239 Pu , 238 U , and 235 U . Estimates are that the total amount of radiation released was about 100 million curies.

Human and medical applications

A person holding a hand held radiation detector near the Chernobyl reactor.
The Chernobyl reactor. More than 100 people died soon after its meltdown, and there will be thousands of deaths from radiation-induced cancer in the future. While the accident was due to a series of human errors, the cleanup efforts were heroic. Most of the immediate fatalities were firefighters and reactor personnel. (credit: Elena Filatova)

What mass of 137 Cs Escaped chernobyl?

It is estimated that the Chernobyl disaster released 6.0 MCi of 137 Cs into the environment. Calculate the mass of 137 Cs released.

Strategy

We can calculate the mass released using Avogadro’s number and the concept of a mole if we can first find the number of nuclei N size 12{N} {} released. Since the activity R size 12{R} {} is given, and the half-life of 137 Cs size 12{"" lSup { size 8{"137"} } "Cs"} {} is found in Appendix B to be 30.2 y, we can use the equation R = 0 . 693 N t 1 / 2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {} to find N size 12{N} {} .

Solution

Solving the equation R = 0 . 693 N t 1 / 2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {} for N size 12{N} {} gives

N = Rt 1/2 0.693 . size 12{N= { { ital "Rt""" lSub { size 8{1/2} } } over {0 "." "693"} } } {}

Entering the given values yields

N = ( 6.0 MCi ) ( 30 . 2 y ) 0 . 693 . size 12{N= { { \( 6 "." 0" MCi" \) \( "30" "." 2" y" \) } over {0 "." "693"} } } {}

Converting curies to becquerels and years to seconds, we get

N = ( 6 . 0 × 10 6 Ci ) ( 3 . 7 × 10 10 Bq/Ci ) ( 30.2 y ) ( 3 . 16 × 10 7 s/y ) 0.693 = 3 . 1 × 10 26 . alignl { stack { size 12{N= { { \( 6 "." 0´"10" rSup { size 8{6} } " Ci" \) \( 3 "." 7´"10" rSup { size 8{"10"} } " Bq/Ci" \) \( "30" "." 2" y" \) \( 3 "." "16"´"10" rSup { size 8{7} } " s/y" \) } over {0 "." "693"} } } {} #" "=3 "." 1´"10" rSup { size 8{"26"} } "." {} } } {}

One mole of a nuclide A X size 12{"" lSup { size 8{A} } X} {} has a mass of A size 12{A} {} grams, so that one mole of 137 Cs size 12{"" lSup { size 8{"137"} } "Cs"} {} has a mass of 137 g. A mole has 6 . 02 × 10 23 size 12{6 "." "02 " times "10" rSup { size 8{"23"} } } {} nuclei. Thus the mass of 137 Cs size 12{"" lSup { size 8{"137"} } "Cs"} {} released was

m = 137 g 6.02 × 10 23 ( 3 . 1 × 10 26 ) = 70 × 10 3 g = 70 kg . alignl { stack { size 12{m= left ( { {"137"" g"} over {6 "." "02 "´"10" rSup { size 8{"23"} } } } right ) \( 3 "." 1´"10" rSup { size 8{"26"} } \) ="70"´"10" rSup { size 8{3} } " g"} {} #" "="70 kg" "." {} } } {}

Discussion

While 70 kg of material may not be a very large mass compared to the amount of fuel in a power plant, it is extremely radioactive, since it only has a 30-year half-life. Six megacuries (6.0 MCi) is an extraordinary amount of activity but is only a fraction of what is produced in nuclear reactors. Similar amounts of the other isotopes were also released at Chernobyl. Although the chances of such a disaster may have seemed small, the consequences were extremely severe, requiring greater caution than was used. More will be said about safe reactor design in the next chapter, but it should be noted that Western reactors have a fundamentally safer design.

Got questions? Get instant answers now!

Activity R size 12{R} {} decreases in time, going to half its original value in one half-life, then to one-fourth its original value in the next half-life, and so on. Since R = 0 . 693 N t 1 / 2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {} , the activity decreases as the number of radioactive nuclei decreases. The equation for R size 12{R} {} as a function of time is found by combining the equations N = N 0 e λt size 12{N=N rSub { size 8{0} } e rSup { size 8{ - λt} } } {} and R = 0 . 693 N t 1 / 2 size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {} , yielding

Questions & Answers

find the equation of the tangent to the curve y=2x³-x²+3x+1 at the points x=1 and x=3
Esther Reply
derivative of logarithms function
Iqra Reply
how to solve this question
sidra
ex 2.1 question no 11
khansa
anyone can help me
khansa
question please
Rasul
ex 2.1 question no. 11
khansa
i cant type here
khansa
Find the derivative of g(x)=−3.
Abdullah Reply
any genius online ? I need help!!
Guzorochi Reply
how can i help you?
Pina
need to learn polynomial
Zakariya
i will teach...
nandu
I'm waiting
Zakariya
plz help me in question
Abish
How can I help you?
Tlou
evaluate the following computation (x³-8/x-2)
Murtala Reply
teach me how to solve the first law of calculus.
Uncle Reply
teach me also how to solve the first law of calculus
Bilson
what is differentiation
Ibrahim Reply
only god knows😂
abdulkadir
f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
Naufal Reply
fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
diron
pweding paturo nsa calculus?
jimmy
how to use fundamental theorem to solve exponential
JULIA Reply
find the bounded area of the parabola y^2=4x and y=16x
Omar Reply
what is absolute value means?
Geo Reply
Chicken nuggets
Hugh
🐔
MM
🐔🦃 nuggets
MM
(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.
Ismael
find integration of loge x
Game Reply
find the volume of a solid about the y-axis, x=0, x=1, y=0, y=7+x^3
Godwin Reply
how does this work
Brad Reply
Can calculus give the answers as same as other methods give in basic classes while solving the numericals?
Cosmos Reply
log tan (x/4+x/2)
Rohan
please answer
Rohan
y=(x^2 + 3x).(eipix)
Claudia
is this a answer
Ismael
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 8

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics for ap® courses' conversation and receive update notifications?

Ask