Question

In all processes where a quantum system decays from a higher energy level to a lower one, the probability that the decay occurs during a given infinitesimal time interval dt is a constant, independent of time. Let us define this fixed probability to be adt, where is some constant (that has nothing to do with wavelength). This expresses the idea that at least for tiny time intervals, the probability is proportional to the duration of the interval: doubling the interval should double the probability. Also let N(1) be the number of undecayed systems at time t. (a) Argue that dN/N = -Adt. (Why the minus sign?) (b) Integrate both sides to show that In(NIN) = -At In particular, explain where the No came from and what this symbol physically means. (c) If we take the exponential of both sides of the above and multiply both sides by No, we get N(t) = Noe-, so a constant probability of decaying during a given fixed infinitesimal interval dt mathematically implies an exponential decay with time in the number of remaining systems in the high-energy (undecayed) state. Now for a radioactive sample, the activity A is the rate at which nuclei decay (that is, the rate that N decreases). Argue that A = AN. (d) Argue that N(t) = Noe-* together with the result above imply that A(t) = Age- Explain how 1 (which physicists call the decay constant) is related to the constants tand 11/2.

Answer 1

# selary Law or delary equation: Let Nt be the number of active nuclides present in a sample at any instant t, then we have, - Ont ant = auta - A No. of 1 is know as the delay Constant. o the negative sige indicates that wt deewares (disintegration) with t. Where (6) 29 o seluanging, the integrate, dire - - xat Nt - lunt = - &t & A (Const) At too, Ne=No, the inittal muber of unclides. A-led. - Rent -- at & ln no & le Nt = – kt. NE Noekt –– (6) By the definition of activity At Ata art) Wt - vorekt dre -Noekt at - At = (- Nr 1 & At a ant NO N= Noe . Tdt 1 ANG

We have, Ni No eat Differentiating wis to t, we obtain, det a-k Noeht - (2) when to (dr) .= - Kwo then from (2) t= 0 Aged de 5) 44 = 40 еке where At Original activity - No. @ from the definition of half-life Ne ne then from (D) No - No é aby a) tye luz0.693 0.693 Again, = the average or imeon value definition + IN. It de - No de we have dra d (see at) = a Noktat - - Ano patekt at on

. te + -htat) 144" I - L { t = 1.443 ty