7.3 Solutions 419
GivenNB/NA= 2. 2 / 0. 53 = 4. 15
λA= 0. 693 /T 1 / 2 (A)= 0. 693 / 12 = 0 .05775 year−^1
λB= 0. 693 /T 1 / 2 (B)= 0. 693 / 18 = 0 .0385 year−^1
We find age of alloyt= 73 .93 years.
7.89 LetWgrams of^210 Po be required.
|dN/dt|=Nλ (1)
Required activity|dN/dt|= 10 × 10 −^3 × 3. 7 × 1010 = 3. 7 × 108 disintegration/
second
N= 6. 02 × 1023 W/ 210 = 2. 867 × 1021 W(2)
λ= 0. 693 /T 1 / 2 = 0. 693 / 138 × 86 , 400 = 5. 812 × 10 −^8 (3)
Use (2) and (3) in (1) and solve forWto obtainW= 2. 22 × 10 −^6 = 2. 22 μg
7.90 Decay constant,λ= 0. 693 / 100 × 86 , 400 = 8 × 10 −^8 s−^1
Let M be the gram molar weight of the substance.
Then number of atomsN=N 0 M= 6. 02 × 1023 M
|dN/dt|=Nλ=N 0 Mλ= 6. 02 × 1023 × 8 × 10 −^8 M= 4. 816 × 1016 M/s
PowerP= 4. 816 × 1016 M× 5 × 10 −^14 =2408 M Watts
But 10% of this power is available, that is 240.8 M Watts. Equating this to
the required power
240. 8 M= 5
OrM= 0 .02 g-molecules
7.91 The present day activity
|dn/dt|=n 0 λe−λt (1)
n 0 = 6. 02 × 1023 × 1. 35 × 10 −^10 / 12 × 100 = 6. 77 × 1010
λ= 3. 92 × 10 −^10 s−^1
dn/dt= 12. 9 / 60 = 0 .215 s−^1
Using these values in (1) and solving fort, we gett= 1. 228 × 1010 sor
390 years
λE λF
7.92 RaE→RaF→RaG
The rate of decay of RaE is given by
dNE/dt=−λENE (1)
whereNEis the number of atoms of RaE andλEits decay constant.
The net change of RaF is given by
dNF/dt=λENE−λFNF (2)
The first term on the right side represents the rate of increase of RaF (Note
the positive sign) and the second term, the rate of decrease (Note the negative
sign). HereNEandNFare the number of atoms of E and F respectively at
timet.