5) The probability density function describing the time (t) between the creation

Question

5) The probability density function describing the time (t) between the creation and decay of a certain unstable elementary particle is given by: f(t)=0 t<0 f(t)=ae^(-?t) t >= 0 with ? and a constant. a) Using the normalization condition find the normalization constant a in terms of ?. b) Find the average time it takes for a particle to decay in terms of ?. c) What is the probability for a particle to "live" more than twice as long as the average time? d) Find the variance of the probability density function in terms of ?

Explanation / Answer

a)

int{f(t) dt} = 1   .......................................... integration from 0 to infinity

==> int{a e^(-lamda t) dt} = 1

==> a/(- lamda) e^(-lamda t) = 1 .......... integration from 0 to infinity

==> a/(- lamda) = 1

==> a = - lamda

b)

t_avrage = int{t f(t) dt}

= int{t a e^(- lamda t) dt}

= int{t (-lamda) e^(- lamda t) dt}

= (-lamda) int{t e^(- lamda t) dt}

= (-lamda) {-((lamda t + 1)*(e^(-lamda t)))/(lamda)^2}

= ((lamda t + 1)*(e^(-lamda t)))/(lamda)

integration from 0 to infinity

==> t_average = 1/(lamda)

c)

integration from "2/lamda" to "infinity":

P = int{f(t) dt}

==> P = int{a e^(-lamda t) dt}

==> P = int{(-lamda) e^(-lamda t) dt}

==> P = e^(- lamda t)

integration from "2/lamda" to "infinity":

==> P = e^(-lamda (2/lamda))

==> P = e^(-2) = 0.135 = 13.5%

c)

variance(t) = int{f(t) (t - (t_average))^2 dt}

variance(t) = int{((-lamda) e^(-lamda t)) (t - (1/lamda))^2 dt}

variance(t) = (e^(-lamda t)) (lamda^2 t^2 + 1)/(lamda^2)

integration from 0 to infinity:

variance(t) = 1/(lamda^2)

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