82
1 Mathematical Physics
Normal equation gives
λ =
t n y n
t 2
n
6
n=1
t n y n = 1 × ln(1/0.835) + 2 × ln(1/0.695)
+ 3 × ln(1/0.58) + 4 × ln(1/0.485)
+ 5 × ln(1/0.405) + 6 × ln(1/0.335)
= 16.5175
6
n=1
t
2
n = 1
2
+ 2
2
+ 3
2
+ 4
2
+ 5
2
+ 6
2
= 91
∴ λ =
16.5175
91
= 0.1815 h
−1
T 1/2 =
0.693
λ
=
0.693
0.1815
= 3.82 h
1.99 We determine the probability P(t) that a given counter records no pulse
during a period t. We divide the interval t into two parts t = t 1 + t 2 . The
probabilities that no pulses are recorded in either the first or the second period
are given by P(t 1 ) and P(t 2 ), respectively, while the probability that no pulse
is recorded in the whole interval is P(t) = P 1 (t 1 + t 2 ).
Since the two events are independent,
P(t 1 + t 2 ) = P(t 1 ) p(t 2 )
The above equation has the solution
P(t) = e
−at
where a is a positive constant. The reason for using the minus sign for a is
that P(t) is expected to decrease with increasing t.
The probability that there will be an event in the time interval dt is c dt. The
combined probability that there will be no events during time interval t, but
one event between time t and t + dt is c e
−at dt where c = constant. It is
readily shown that c = a. This follows from the normalization condition
∞
0
P(t)dt = c
∞
0
e
−at dt = 1
Thus d p(t) = ae
−at dt
Clearly, small time intervals are more favoured than large time intervals
amongst randomly distributed events. If the data have large number N of
intervals then the number of intervals greater than t 1 but less than t 2 is
n = N
t2
t 1
ae
−at dt = N (e
−at 1 − e
−at 2 ) (Interval distribution)
a represents the average number of events per unit time.