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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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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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