Consider the alpha particles emitted by a radioactive source during some time interval t.?

Question

Let t to be subdivided into many small intervals of length delta t.Since particles are emitted at random times, the probability of a radioactive disintegration occurring during any delta t is completely independent of whatever disintegrations occur at other times.Let delta t be small enough so that the probability of more than 1 disintegration occurring in delta t is negligibly small.This means that there is some probability p of 1 disintegration occurring during a time delta t (with p<<1)& probability 1-p of no disintegration occurring during this time.Each such time interval delta t can then be regarded as an independent trial,there being a total of N=t/delta t such trials during a time t. A)Show that the probability W(n) of n dsintegratns occurring in t is given by Poisson distrbtn.B)If the strength of the radioactive source is such that mean # of disintegrations/minute=24.What is the probblty of getting n counts in 10 secs?Get num values for all integral values of n from 0 to 8.

Answer 1

Small fix: your probability p that 1 disintegration occurs within dt (=delta t) should be equal p*dt, i.e. proportional with dt. p is usually marked by lambda.
Now divide t into M*dt. Then W(n) = (M over n)*((p*dt)^n)* (1-p*dt)^(M-n) = (M*(M-1)*(M-2)*...*(M-n+1)/n!)*((p*t)^n/M^n)*((1-p*t/M)^(M-n)). When we take a limit when M goes infinite we have:
M*(M-1)*(M-2)*...*(M-n+1)/(M^n) = 1
(1 - pt/M)^M = e^(-pt)
(1-pt/M)^n = 1
Therefore:
A) W(n) = ((p*t)^n/n!)*e(-pt) which is Poisson distribution probability of having n disintegration within time t.
B) From A also follows that average number of disintegration per unit of time = p. Therefore average number of disintegration per second = 24/60 = 0.4. For 10 sec interval p*t = 4. Therefore for n=0,1,2,3,4,5,6,7,8 W(n) = 0.018316, 0.073263, 0.146525, 0.195367, 0.195367, 0.156293, 0.104196, 0.02977.

Answer 2

The question is a sprint ambiguous. "Lifetime" can advise a minimum of two various issues. A 0.5-existence is the time is takes for 0.5 the stuff to decay. A function lifetime (tau) is the time it takes so as that in basic terms a fragment a million/e of the stuff continues to be. My terrific wager would be that the 624 years is a function lifetime. So in case you opt on 3/4 (quantity necessary / commencing quantity) of the stuff to proceed to be, meaning that: a million/e ^ (t / tau) = (quantity necessary / commencing quantity) t = tau ln (commencing quantity / quantity necessary) just to respond to your question, if my assumption is genuine, the 0.5-existence would be tau over organic log of two.