1000 Solved Problems in Modern Physics

8.3 Solutions 479

We can now calculatep, the resonance escape probability.

Σs/N 0 =

Σs(U)+Σs(m)

N 0

=σs(U)+

Nm

Nu

σa(M)

= 8. 3 + 402. 9 × 4. 8 =1942 b

We can use the empirical relation for the effective resonance integral (ERI)

∫E^0

E

(σa)eff

dE

E

= 3. 85 (Σs/N 0 )^0.^415 = 3 .85(1942)^0.^145 =89 b

We have ignored the contribution of Uranium to the scattering as its inclu-

sion hardly changes the result. ThusΣ 0 /N 0  Σs/N 0 =

Nm

N 0

σs(M) =

402. 9 × 4. 8 = 1934

p=exp

[

−(ERI)

/

Σ 0 ξ

N 0

]

=exp

(

−

89

1934 × 0. 158

)

= 0. 747

The reproduction factor

k∞=ξnfp=(1.0)(1.34)(0.857)(0.747)= 0. 858

Thus,k∞<1, and so the reactor cannot go critical.

8.87 The spatial distribution was derived for Problem 8.78.

φ(r)=

3 Q

4 πλtr

e−r/L

r

(1)

If 1% of the neutrons are to escape then

φ(r)

Q

=

1

100

=

3

4 πλtr

e−r/L

r

(2)

L=

(

λtrλa

3

) 1 / 2

λs=

1

Σs

=

A

σsN 0 ρ

=

9

5. 6 × 10 −^24 × 6. 02 × 1023 × 1. 85

= 1 .443 cm

λs=

1

Σa

=

A

σaN 0 ρ

=

9

10 × 10 −^27 × 6. 02 × 1023 × 1. 85

=808 cm

λtr=

λs

1 − 32 A

=

1. 443

1 −

2

3 × 9

= 1 .564 cm

L=

(

1. 564 × 808

3

) 1 / 2

= 20 .52 cm

Inserting the values ofλtrandLin (2) and solving forr, we getr= 9 .6cm.

Thus the radius ought to be greater than 9.6 cm.