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R = R 0 e λt , size 12{R=R rSub { size 8{0} } e rSup { size 8{ - λt} } } {}

where R 0 size 12{R rSub { size 8{0} } } {} is the activity at t = 0 size 12{t=0} {} . This equation shows exponential decay of radioactive nuclei. For example, if a source originally has a 1.00-mCi activity, it declines to 0.500 mCi in one half-life, to 0.250 mCi in two half-lives, to 0.125 mCi in three half-lives, and so on. For times other than whole half-lives, the equation R = R 0 e λt size 12{R=R rSub { size 8{0} } e rSup { size 8{ - λt} } } {} must be used to find R size 12{R} {} .

Phet explorations: alpha decay

Watch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half life.

Alpha Decay

Test prep for ap courses

A radioactive sample has N atoms initially. After 3 half-lives have elapsed, how many atoms remain?

  1. N/3
  2. N/6
  3. N/8
  4. N/27
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When P 84 215 o MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaiIdacaaI0aaabaGaaGOmaiaaigdacaaI1aaaaOGaaeiuaiaab+gaaaa@3BA7@ decays, the product is P 82 211 b. MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0raaSqaaiaaiIdacaaIYaaabaGaaGOmaiaaigdacaaIXaaaaOGaaeiuaiaabkgaaaa@3B94@ The half-life of this decay process is 1.78 ms. If the initial sample contains 3.4 x 10 17 parent nuclei, how many are remaining after 35 ms have elapsed? What kind of decay process is this (alpha, beta, or gamma)?

This must be alpha decay since 4 nucleons (2 positive charges) are lost from the parent nucleus. The number remaining is found from:

N ( t ) = N 0 e ( 0.693 t t 1 2 ) = 3.4 × 10 17 e ( ( 0.693 ) ( 0.035 ) 0.00173 ) MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6233@

N ( t ) = 4.1 × 10 11 MathType@MTEF@5@5@+=feaagyart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeOtamaabmaabaGaamiDaaGaayjkaiaawMcaaiabg2da9iaaisdacaGGUaGaaGymaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacaaIXaGaaGymaaaaaaa@41A6@ nuclei

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Section summary

  • Half-life t 1 / 2 size 12{t rSub { size 8{1/2} } } {} is the time in which there is a 50% chance that a nucleus will decay. The number of nuclei N size 12{N} {} as a function of time is
    N = N 0 e λt , size 12{N=N rSub { size 8{0} } e rSup { size 8{ - λt} } } {}
    where N 0 size 12{N rSub { size 8{0} } } {} is the number present at t = 0 size 12{t=0} {} , and λ size 12{λ} {} is the decay constant, related to the half-life by
    λ = 0 . 693 t 1 / 2 . size 12{λ= { {0 "." "693"} over {t rSub { size 8{1/2} } } } } {}
  • One of the applications of radioactive decay is radioactive dating, in which the age of a material is determined by the amount of radioactive decay that occurs. The rate of decay is called the activity R size 12{R} {} :
    R = Δ N Δ t . size 12{R= { {ΔN} over {Δt} } } {}
  • The SI unit for R size 12{R} {} is the becquerel (Bq), defined by
    1 Bq = 1 decay/s. size 12{1" Bq"="1 decay/s"} {}
  • R size 12{R} {} is also expressed in terms of curies (Ci), where
    1 Ci = 3 . 70 × 10 10 Bq. size 12{1" Ci"=3 "." "70" times "10" rSup { size 8{"10"} } " Bq"} {}
  • The activity R size 12{R} {} of a source is related to N size 12{N} {} and t 1 / 2 size 12{t rSub { size 8{1/2} } } {} by
    R = 0 . 693 N t 1 / 2 . size 12{R= { {0 "." "693"N} over {t rSub { size 8{1/2} } } } } {}
  • Since N size 12{N} {} has an exponential behavior as in the equation N = N 0 e λt size 12{N=N rSub { size 8{0} } e rSup { size 8{ - λt} } } {} , the activity also has an exponential behavior, given by
    R = R 0 e λt , size 12{R=R rSub { size 8{0} } e rSup { size 8{ - λt} } } {}
    where R 0 size 12{R rSub { size 8{0} } } {} is the activity at t = 0 size 12{t=0} {} .

Conceptual questions

In a 3 × 10 9 size 12{3 times "10" rSup { size 8{9} } } {} -year-old rock that originally contained some 238 U , which has a half-life of 4.5 × 10 9 years, we expect to find some 238 U remaining in it. Why are 226 Ra , 222 Rn , and 210 Po also found in such a rock, even though they have much shorter half-lives (1600 years, 3.8 days, and 138 days, respectively)?

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Does the number of radioactive nuclei in a sample decrease to exactly half its original value in one half-life? Explain in terms of the statistical nature of radioactive decay.

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Radioactivity depends on the nucleus and not the atom or its chemical state. Why, then, is one kilogram of uranium more radioactive than one kilogram of uranium hexafluoride?

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Explain how a bound system can have less mass than its components. Why is this not observed classically, say for a building made of bricks?

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Spontaneous radioactive decay occurs only when the decay products have less mass than the parent, and it tends to produce a daughter that is more stable than the parent. Explain how this is related to the fact that more tightly bound nuclei are more stable. (Consider the binding energy per nucleon.)

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Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
what is the solution
bill
how did you arrive at this answer?
bill
-24m+3+3mÁ^2
Susan
i really want to learn
Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
Hw did u arrive to this answer.
Aphelele
hi
Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
Hi
Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
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how far
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cool u
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state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
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hello
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Method
I am eliacin, I need your help in maths
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how can I help
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Amoon
however, may I ask you some questions about Algarba?
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hi
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what the last part of the problem mean?
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The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
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Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
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difference between rational and irrational numbers
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When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
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Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
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d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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