Simple Nature - Light and Matter

13.1.5 Applications of calculus

The area under the probability distribution is of course an in-

tegral. If we call the random numberxand the probability distri-

butionD(x), then the probability thatxlies in a certain range is

given by

(probability ofa≤x≤b) =

∫b

a

D(x) dx.

What about averages? Ifxhad a finite number of equally probable

values, we would simply add them up and divide by how many we

had. If they weren’t equally likely, we’d make the weighted average

x 1 P 1 +x 2 P 2 +... But we need to generalize this to a variablexthat

can take on any of a continuum of values. The continuous version

of a sum is an integral, so the average is

(average value ofx) =

∫

xD(x) dx,

where the integral is over all possible values ofx.

Probability distribution for radioactive decay example 5

Here is a rigorous justification for the statement in subsection

13.1.4 that the probability distribution for radioactive decay is found

by substitutingN(0) = 1 into the equation for the rate of decay. We

know that the probability distribution must be of the form

D(t) =k0.5t/t^1 /^2 ,

wherekis a constant that we need to determine. The atom is

guaranteed to decay eventually, so normalization gives us

(probability of 0≤t<∞) = 1

=

∫∞

0

D(t) dt.

The integral is most easily evaluated by converting the function

into an exponential witheas the base

D(t) =kexp

[

ln

(

0.5t/t^1 /^2

)]

=kexp

[

t

t 1 / 2

ln 0.5

]

=kexp

(

−

ln 2

t 1 / 2

t

)

,

which gives an integral of the familiar form

∫

ecxdx = (1/c)ecx.

We thus have

1 =−

k t 1 / 2

ln 2

exp

(

−

ln 2

t 1 / 2

t

)]∞

0

,

which gives the desired result:

k=

ln 2

t 1 / 2

.

868 Chapter 13 Quantum Physics