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Extracts from this document...

Introduction

1st consider an analogous model.

In the dice analogy of radioactive decay, the number of dice which decayed per throw (∆N) is given by

∆N = N

∆ means a small change

N is the total number of dice at start

The minus sign indicated that ∆N is ever decreasing and the comes from the probability that advice will show a six when thrown.

2nd – Now consider a real example of radioactive decay. In any such example, if k is the change of decay per second then in a time interval ∆t, a fraction k∆t will decay.

The “k∆t” term here is equivalent to the

Middle

dt

Where N          = number of atoms at time t=0

Where N            = number of atoms at time t=t

=

=        -kt

N = N. e

Half life

This is the time taken for N to become.

i.e. the number of atoms remaining after time t to become equal to half of the original number of atoms present.

Conclusion

ity/15668/html/images/image11.png" style="width:41.33px;height:41.33px;margin-left:0px;margin-top:0px;" alt="image11.png" />e

Taking log to the base e gives us:

t =

The activity of a radioactive material

This is exactly the same thing as its rate of decay. Radioactive sources are marked with “strengths” or activities in curies 1 Curie = 3.7*10 disintegrations per second. OR nowadays in Bequerels 1Bq = 1 disintegration per second.

Since activity = rate of decay.

A =

And since

N = Ne

A =

A = -kNe

But

N = Ne

A = -kN

By the use of substitution we come get the following equation.

A = Ae

This student written piece of work is one of many that can be found in our GCSE Radioactivity section.

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