1.3 Solutions 83
Two limiting cases
(a)t 2 =∞. We find that the number of intervals greater than any duration is
Ne−atin whichat=average number of events in time t. In the case
of radioactivity at timet=0, letN=N 0.
Then the radioactive decay law becomes
N=N 0 e−λt
whereNis the number of surviving atoms at timet, anda=λis the
decay constant, that is the number of decays per unit time.
(b)t 1 =0, implies that the number of events shorter than any durationtis
N 0 (1−e−at)
For radioactive decay the above equation would read for the number of
decays in time interval 0 tot.
N=N 0 (1−e−λt)1.100 Ns=N 0 −NB=^14.^5 −^10 =^4.^5
σs=√
10
t+
14. 5
t=
√
24. 5
t
σs
Ns=
5
100
=
1
4. 5
√
24. 5
t
t=484 min1.101 The best values ofa 0 ,a 1 anda 2 are found by the Least square fit. The residue
Sis given by
S=∑ 6
n= 1 (yn−a^0 −a^1 xn−a^2 x2
n)
2Minimize the residue.
∂S
∂a 0= 0 ,gives∑ 6
n= 1 yn=na^0 +a^1∑
xn+a 2∑
xn^2 (1)∂S
∂a 1=0gives
∑
xnyn=a 0∑
xn+a 1∑
xn^2 +a 2∑
xn^3 (2)∂S
∂a 2=0gives
∑
x^2 nyn=a 0∑
xn^2 +a 1∑
xn^3 +a 2∑
x^4 n (3)Equations (1), (2) and (3) are the so-called normal equations which are to be
solved as ordinary simultaneous equations.