1000 Solved Problems in Modern Physics

82 1 Mathematical Physics

Normal equation gives

λ=

∑

tnyn

∑

6 t^2 n

∑

n= 1

tnyn= 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

tn^2 = 12 + 22 + 32 + 42 + 52 + 62 = 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 probabilityP(t) that a given counter records no pulse

during a periodt. We divide the intervaltinto two partst=t 1 +t 2 .The

probabilities that no pulses are recorded in either the first or the second period

are given byP(t 1 ) andP(t 2 ), respectively, while the probability that no pulse

is recorded in the whole interval isP(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

whereais a positive constant. The reason for using the minus sign forais

thatP(t) is expected to decrease with increasingt.

The probability that there will be an event in the time interval dtiscdt.The

combined probability that there will be no events during time intervalt,but

one event between timetandt+dtisce−atdtwherec=constant. It is

readily shown thatc=a. This follows from the normalization condition

∫∞

0

P(t)dt=c

∫∞

0

e−atdt= 1

Thus dp(t)=ae−atdt

Clearly, small time intervals are more favoured than large time intervals

amongst randomly distributed events. If the data have large numberNof

intervals then the number of intervals greater thant 1 but less thant 2 is

n=N

∫t 2

t 1

ae−atdt=N(e−at^1 −e−at^2 ) (Interval distribution)

arepresents the average number of events per unit time.