Average Life of a Radioactive element

Average life of a radioactive element is given by,

τ = 1/λ

Example : Nuclei of a radio-active element A are being produced at a constant rate α. The element has a decay constant λ. At time t = 0, there are no nuclei of the element present.

(a) Calculate the number of nuclei A as a function of time t.

(b) If α = 2N0 λ , calculate the number of nuclei of A after one-half of A and also the limiting value of N as t →∞.

Solution:

(a) X → A → B

Net rate of formulation of A is

$ \displaystyle \frac{dN}{dt} = \alpha – \lambda N $

$ \displaystyle \int \frac{dN}{\alpha – \lambda N} = \int dt $

$ \displaystyle -\frac{1}{\lambda}ln(\alpha – \lambda N) = t + C $

Where C is the constant of integration

at t = 0 , N = No ;

$ \displaystyle C = -\frac{1}{\lambda}ln(\alpha – \lambda N_0) $

$ \displaystyle t = \frac{1}{\lambda}ln (\frac{\alpha – \lambda N_0}{\alpha – \lambda N}) $

$ \displaystyle N = \frac{\alpha -(\alpha – \lambda N_0)e^{-\lambda t}}{\lambda} $

(b) When α = 2 N0 λ & t = T1/2 = ln2/λ

$ \displaystyle N = \frac{2 N_0 \lambda -(2 N_0 \lambda – \lambda N_0)e^{-ln2}}{\lambda} = \frac{3N_0}{2}$

also t → ∞ ; N = 2No

Exercise : An experiment is performed to determine the half-life of a radioactive substance which emits one beta particle for each decay process. Observations show that an average of 8.4 β -particles are emitted each second by 2.5 mg of the substance. The atomic weight of the substance is 230. Calculate the half-life of the substance.

Example: It is proposed to use the nuclear fusion reaction

1H2 + 1H22He4

in a nuclear reactor with an electrical power rating of 200 MW. If the energy from the above reaction is used with 25 percent efficiency in the reactor, how many grams of deuterium fuel will be needed per day ? (The masses of 1H2 and 2He4 are 2.0141 amu and 4.0026 amu respectively.)

Solution: Mass defect occurring in one fusion reaction

$ \displaystyle \Delta m = 2\times 2.0141 – 4.0026 $

= 0.0256 amu

Energy released = 0.0256 × 931 MeV

Energy used in reactor per fusion reaction

= (25/100) × 0.0256 × 931 MeV

= 5.9584 MeV

= 9.5334 x 10-13 J

Total energy required per day = (200 MW) × (24 × 60 × 60 sec.)

Mass of deuterium fuel needed per reaction = 2 × 2.0141 amu

= 4.082/(6.02×1023)

= 0.6691 × 10-23 gm

Mass of deuterium required

$ \displaystyle = \frac{0.6691 \times 10^{-23}\times 200 \times 10^6 \times 24 \times 60 \times 60}{9.5334\times 10^{-13}} $

= 120 gm.

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