1000 Solved Problems in Modern Physics

(Grace) #1

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 min

1.101 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 xn^2 )^2

Minimize 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


x^2 n+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.
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