1.3 Solutions
83
Two limiting cases
(a) t 2 = ∞. We find that the number of intervals greater than any duration is
N e
−at in which at = average number of events in time t. In the case
of radioactivity at time t = 0, let N = N 0 .
Then the radioactive decay law becomes
N = N 0 e
−λt
where N is the number of surviving atoms at time t, and a = λ is 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 duration t is
N 0 (1 − e
−at )
For radioactive decay the above equation would read for the number of
decays in time interval 0 to t.
N = N 0 (1 − e
−λt )
1.100 N s = N 0 − N B = 14.5 − 10 = 4.5
σ s =
10
t
+
14.5
t
=
24.5
t
σ s
N s
=
5
100
=
1
4.5
24.5
t
t = 484 min
1.101 The best values of a 0 , a 1 and a 2 are found by the Least square fit. The residue
S is given by
S =
6
n=1
(y n − a 0 − a 1 x n − a 2 x
2
n )
2
Minimize the residue.
∂ S
∂a 0
= 0, gives
6
n=1
y n = na 0 + a 1
x n + a 2
x
2
n
(1)
∂ S
∂a 1
= 0 gives
x n y n = a 0
x n + a 1
x
2
n + a 2
x
3
n
(2)
∂ S
∂a 2
= 0 gives
x
2
n y n = a 0
x
2
n + a 1
x
3
n + 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.