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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

If a prism is fully imersed in water then the ray of light will normally dispersed or their is any difference?
Anurag Reply
the same behavior thru the prism out or in water bud abbot
If this will experimented with a hollow(vaccum) prism in water then what will be result ?
What was the previous far point of a patient who had laser correction that reduced the power of her eye by 7.00 D, producing a normal distant vision power of 50.0 D for her?
Jaydie Reply
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
29/20 ? maybes
In what ways does physics affect the society both positively or negatively
Princewill Reply
how can I read physics...am finding it difficult to understand...pls help
rerry Reply
try to read several books on phy don't just rely one. some authors explain better than other.
And don't forget to check out YouTube videos on the subject. Videos offer a different visual way to learn easier.
hope that helps
I have a exam on 12 february
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what is velocity
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It's Reply
what happens to the size of charge if the dielectric is changed?
Brhanu Reply
omega= omega not +alpha t derivation
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u have to derivate it respected to time ...and as w is the angular velocity uu will relace it with "thita × time""
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Saeed Reply
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Ali Reply
displacement in easy way.
Mubashir Reply
binding energy per nucleon
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Practice Key Terms 8

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