1000 Solved Problems in Modern Physics

(Romina) #1

514 9 Particle Physics – I


9.41 Slope=


2%

100 V

=

0 .1%

5 V

The voltage should not vary more than±5 V from the operating voltage of
800 V.

9.42 For the cylindrical geometry of the G.M. tube the electric field is given by
E=V/rln(b/a)
whereVis the applied voltage,ris the distance of a point from the anode,b
andaare the diameters of the cathode and the anode wire.
The field will be maximum close to the anode.
r= 0 .1mm= 10 −^4 m
Emax= 1 , 000 / 10 −^4 ln(20/ 0 .2)= 2. 17 × 106 V/m
The lifetime of G.M. tube in years is given by dividing total number of
possible counts by counts per year.
t= 109 /(52× 30 × 60 × 3 ,000)= 3 .56 years.
In practice, the G.M. tube will not work properly long before the above esti-
mate.


9.43 As the mean free path is inversely proportional to the pressure, the mean free
path at 10 cm pressure will be 2× 10 −^4 ×(76/10) or 1. 52 × 10 −^3 cm.
At a distancer =mean free path= 1. 52 × 10 −^3 cm. from the anode,
the electric field should be such that the electron acquires sufficient energy to
ionize Argon for which the ionization energy is 15.7 eV. The required value of
Eis
E= 15. 7 / 1. 52 × 10 −^3 = 1. 03 × 104 V/cm
r=V/Eln(b/a)= 1000 / 1. 03 × 104 ln(20/ 0 .1)= 0 .0183 cm= 0 .183 mm


9.44 The source gives 3. 7 × 107 × 25 = 9. 25 × 108 disintegrations/second
G.M. Counting rate of beta rays plus gamma rays
=(2, 000 /60)−(750/300)= 30. 83 /s
Therefore, efficiency= 30. 83 /(9. 25 × 108 )= 3. 33 × 10 −^8
Scintillation counter counting rate of beta particles=(9, 300 /60)−(300/300)
= 154 /s
Therefore, efficiency= 154 /(9. 25 × 108 )= 16. 65 × 10 −^8
For both the counters the efficiency is low because of small solid angle of
acceptance. Both the counters would register beta particles as well as gamma
rays. But the efficiency for counting gamma rays in G.M. counter will be quite
low, being of the order of 1%. This is because gamma rays cause ionization
only indirectly by hitting the walls of the GM counter and ejecting electrons.
The efficiency is relatively higher in the scintillation counter. For this reason,
in the given situation the efficiency for scintillation counter is approximately
five times greater.


9.45 The true counting rate n is related to the observed counting raten 0 by
n=n 0 /(1−n 0 τ)

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