1000 Solved Problems in Modern Physics

9.3 Solutions 515

whereτis the dead time.τ= 100 μs= 10 −^4 s.

n 0 = 104 /min= 166. 7 /s

n=(10^4 /min)/(1− 166. 7 × 10 −^4 )= 10 , 169 /min

9.46 V=V 0 e−t/RC
et/RC=V 0 /V= 100 / 99
R=t/Cln(100/99)= 86 , 400 /(5× 10 −^12 )(0.01)= 1. 73 × 1018 ohm.

9.47 V=Erln(d^1 /d^2 )(1)
Putr=d 2 / 2 = 0. 01 / 2 = 0 .005 cm
E= 15. 7 /λ= 15. 7 / 8 × 10 −^4 V/cm= 19 ,625 V/cm
d 1 /d 2 = 4 / 0. 01 = 400
V=(19,625)(0.005) ln 400=588 V.

9.48 Q=CV
Initial chamber charge:
Qch=^5 ×^10 −^12 ×^600 =^3.^0 ×^10 −^9 Coulomb
Initial electrometer charge:
Qel= 0. 4 × 10 −^12 × 600 = 0. 24 × 10 −^9 Coulomb
Chamber charge lost:
8 × 109 × 1. 6 × 10 −^19 = 1. 28 × 10 −^9 Coulomb
Final chamber charge:
Qch′(3. 0 − 1 .28)× 10 −^9 = 1. 72 × 10 −^9
Final electrometer charge:
Q′el= 0. 24 × 10 −^19 Coulomb
Final system charge
Q=(1. 72 + 0 .24)× 10 −^9 = 1. 96 × 10 −^9 Coulomb
Total capacitanceC=(5. 0 + 0 .4)× 10 −^12 F=(5. 4 × 10 −^12 )
Final potential= 1. 96 × 10 −^9 / 5. 4 × 10 −^12 = 363 V.

9.3.6 Scintillation Counter...............................

9.49 The velocity of a particle
β=p/E=p/(p^2 +m^2 )^1 /^2
The time taken to cross a distanceLis
t=L/βc=(L/c)[(p^2 +m^2 )/p^2 ]^1 /^2 =(L/c)[(1+(m^2 /p^2 )]^1 /^2
The difference in time taken for proton and pion is
Δt=tp−tπ=(L/c)

{[

1 +

(

mp^2 /p^2

)] 1 / 2

−

[

1 +

(

mπ^2 /p^2

)] 1 / 2 }

Substituting,Δt= 100 × 10 −^12 s,c= 3 × 108 ms−^1

mp= 0 .94 GeV/c^2 ,mπ= 0 .14 GeV/c^2 andp=4GeV/c, and solving for

L, we find the minimum value ofL= 1 .14 m

9.50t=

d

c

(

1 +

m^2

p^2

) 1 / 2

Putt= 51 × 10 −^9 s,c= 3 × 108 m/s,d=12 m andp= 1 .19 GeV/c and

solve formto findmp−= 0 .94 GeV/c^2 for the anti proton mass.