By mass spectral analysis, a sample of strontium is known to contain 2.64times10

Question

By mass spectral analysis, a sample of strontium is known to contain 2.64times10^10 atoms of Sr-90 as the only radioactive element. The absolute disintegration rate of this sample is measured as 1238 disintegrations per minute. Calculate the half-life (in years) of Sr-90. How long will it take (in years) for the disintegration rate of this sample to drop to 829 disintegrations per minute? In a living organism, the decay of C-14 produces 15.3 disintegrations per minute per gram of carbon. The half-life of C-14 is 5730 years. A bone sample with 2.5 g of carbon has 19.8 disintegrations per minute. How old is the bone sample in years? When^239Pu is used in a nuclear reactor, one of the fission events that occurs is The atomic mass of each atom is given below its symbol in the equation. Find the energy released (in kJ) when 4.50 g of plutonium undergoes this particular fission. (1) Calculate the change in mass, expressed in atomic mass units, that accompanies this fission event. (2) Convert the mass loss to joules per fission. (3) Find the energy released per mole of plutonium... One of the fission products that causes major concern is Sr-90, since it is incorporated into milk and other high-calcium foods. Sr-90 undergoes beta decay with a half-life of 28.1 years. What percentage of the Sr-90 that was formed in the detonation of the first fission bombs in August 1945 is still present in the environment in August 2027?

Explanation / Answer

29)

A = k x N

A = activity

k = decay constant

N = number of atoms

so

1238 = k x 2.64 x 10^10

k = 4.69 x 10-8

now

half life = ln2 / k

half life = ln2 / 4.69 x 10-8

half life = 1.478 x 10^7 min

1 yr = 525600 min

so

half life = 1.478 x 10^7 / 525600

half life = 28.12 years

30)

now

ln (A/Ao) = -k x t

so

ln ( 829 / 1238) = - 4.69 x 10-8 x t

t = 8.55 x 10^6 min

t = 8.55 x 10^6 / 525600

t = 16.27

so

it takes 16.27 years

31)

we know that

decay constant (k) = ln2 / half life

k = ln2 / 5730

k = 1.21 x 10-4

now

A = disintegrations per minute per gram = 19.8 / 2.5 = 7.92

now

ln (A/Ao) = -kt

ln ( 7.92 / 15.3) = - 1.21 x 10-4 x t

t = 5443.266

so

bone sample is 5443.266 years old

35)

mass defect (dm) = mass of reactants - mass of products

mass defect = ( 1.008665u + 239.052u) - ( 95.916u + 139.917u + 4 x1.008665u)

dm = 0.193005u

now

dE = dm x 931.5 MeV

dE = 0.193005 x 931.5

dE = 179.784 MeV

now

1 MeV = 1.6 x 10-13 J

so

dE = 179.784 x 1.6 x 10-13 J

dE = 2.876 x 10-11 J

dE = 2.876 x 10-14 kJ

this is for one atom of plutonium

now

number of atoms fo Plutonium = mass x 6.023 x 10^23 / atomic mass

number of atoms of plutonium = 1.1338 x 10^22

now

total energy released = 1.1338 x 10^22 x 2.876 x 10-14

total energy released = 3.26 x 10^8 kJ

so

3.26 x 10^8 kJ of energy is released

36)

decay constant (k) = ln2 / half life

decay constant (k) = ln2 / 28.1 = 0.024667

now

ln ( N/No) = -kt

ln (N/No) = - 0.024667 x 82

N/No = 0.1323

(N/No) x 100 = 0.1323 x 100

(N/No) x 100 = 13.23

so

13.23% still remains

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.