1000 Solved Problems in Modern Physics

412 7 Nuclear Physics – I

7.67 Number of gamma rays absorbed in the thicknessxcm of lead
N=N 0 (1−e−μx)
whereN 0 is the initial number andμis the absorption coefficient expressed in
cm−^1.
Nowμ=σN 0 ρ/A
whereN 0 is the Avagadro’s number,ρis the density of lead andAis its atomic
weight.
μ= 20 × 10 −^24 × 6. 02 × 1023 × 11. 3 / 207 = 0 .657 cm−^1
n/n 0 = 90 / 100 = 1 −e−^0.^657 x
x= 3 .5cm

7.68 The K-shell absorption wavelength in Ag
λk= 0 .0424 nm
The corresponding energy
EK(Ag)= 1 , 241 / 0. 0485 = 25 ,567 eV
We use the formula
EK(Z)= 13 .6(Z−σ)^2
For silver 25. 567 × 103 = 13 .6(47−σ)^2
whenceσ= 3. 64
For the impurityXof atomic numberZ= 50
Ek(Z)= 13 .6(50− 3 .64)^2
= 29 ,229 eV
= 29 .23 keV
Now the wavelength of 0.0424 nm corresponds toE= 29 .24 keV which is
in agreement with the calculated value. Thus the impurity is 50 Sn

7.69eV=hc/λ−W
AplotofVagainst 1/λmust be a straight line. The slope of the line gives
hc/e, hencehcan be determined. The intercept multiplied byhcgiveW,the
work function. The threshold frequency is given byν 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