412
7 Nuclear Physics – I
7.67 Number of gamma rays absorbed in the thickness x cm of lead
N = N 0 (1 − e
−μ x )
where N 0 is the initial number and μ is the absorption coefficient expressed in
cm
−1 .
Now μ = σ N 0 ρ/A
where N 0 is the Avagadro’s number, ρ is the density of lead and A is its atomic
weight.
μ = 20 × 10
−24
× 6.02 × 10
23
× 11.3/207 = 0.657 cm
−1
n/n 0 = 90/100 = 1 − e
−0.657x
x = 3.5 cm
7.68 The K-shell absorption wavelength in Ag
λ k = 0.0424 nm
The corresponding energy
E K (Ag) = 1,241/0.0485 = 25,567 eV
We use the formula
E K (Z ) = 13.6(Z − σ )
2
For silver 25.567 × 10
3
= 13.6(47 − σ )
2
whence σ = 3.64
For the impurity X of atomic number Z = 50
E k (Z ) = 13.6(50 − 3.64)
2
= 29,229 eV
= 29.23 keV
Now the wavelength of 0.0424 nm corresponds to E = 29.24 keV which is
in agreement with the calculated value. Thus the impurity is 50 Sn
7.69 eV = hc/λ − W
A plot of V against 1/λ must be a straight line. The slope of the line gives
hc/e, hence h can be determined. The intercept multiplied by hc give W , the
work function. The threshold frequency is given by ν 0 = W/ h (Fig. 7.16).
Fig. 7.16
0.0
0.5
1.0
1.5
0.0
1.0
2.0
3.0
V