7.3 Solutions
419
Given N B /N A = 2.2/0.53 = 4.15
λ A = 0.693/T 1/2 (A) = 0.693/12 = 0.05775 year
−1
λ B = 0.693/T 1/2 (B) = 0.693/18 = 0.0385 year
−1
We find age of alloy t = 73.93 years.
7.89 Let W grams of
210 Po be required.
|dN /dt| = N λ
(1)
Required activity |dN /dt| = 10×10
−3
×3.7×10
10
= 3.7×10
8 disintegration/
second
N = 6.02 × 10
23 W/210 = 2.867 × 10
21 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 for W to obtain W = 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 atoms N = N 0 M = 6.02 × 10
23 M
|dN /dt| = N λ = N 0 Mλ = 6.02 × 10
23
× 8 × 10
−8 M = 4.816 × 10
16 M/s
Power P = 4.816 × 10
16 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
Or M = 0.02 g-molecules
7.91 The present day activity
|dn/dt| = n 0 λ e
−λt
(1)
n 0 = 6.02 × 10
23
× 1.35 × 10
−10
/12 × 100 = 6.77 × 10
10
λ = 3.92 × 10
−10 s
−1
dn/dt = 12.9/60 = 0.215 s
−1
Using these values in (1) and solving for t, we get t = 1.228 × 10
10 s or
390 years
λ E λ F
7.92 RaE → RaF → RaG
The rate of decay of RaE is given by
dN E /dt = −λ E N E
(1)
where N E is the number of atoms of RaE and λ E its decay constant.
The net change of RaF is given by
dN F /dt = λ E N E − λ F N F
(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). Here N E and N F are the number of atoms of E and F respectively at
time t.
419
Given N B /N A = 2.2/0.53 = 4.15
λ A = 0.693/T 1/2 (A) = 0.693/12 = 0.05775 year
−1
λ B = 0.693/T 1/2 (B) = 0.693/18 = 0.0385 year
−1
We find age of alloy t = 73.93 years.
7.89 Let W grams of
210 Po be required.
|dN /dt| = N λ
(1)
Required activity |dN /dt| = 10×10
−3
×3.7×10
10
= 3.7×10
8 disintegration/
second
N = 6.02 × 10
23 W/210 = 2.867 × 10
21 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 for W to obtain W = 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 atoms N = N 0 M = 6.02 × 10
23 M
|dN /dt| = N λ = N 0 Mλ = 6.02 × 10
23
× 8 × 10
−8 M = 4.816 × 10
16 M/s
Power P = 4.816 × 10
16 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
Or M = 0.02 g-molecules
7.91 The present day activity
|dn/dt| = n 0 λ e
−λt
(1)
n 0 = 6.02 × 10
23
× 1.35 × 10
−10
/12 × 100 = 6.77 × 10
10
λ = 3.92 × 10
−10 s
−1
dn/dt = 12.9/60 = 0.215 s
−1
Using these values in (1) and solving for t, we get t = 1.228 × 10
10 s or
390 years
λ E λ F
7.92 RaE → RaF → RaG
The rate of decay of RaE is given by
dN E /dt = −λ E N E
(1)
where N E is the number of atoms of RaE and λ E its decay constant.
The net change of RaF is given by
dN F /dt = λ E N E − λ F N F
(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). Here N E and N F are the number of atoms of E and F respectively at
time t.