Nuclear Physics Concept Page - 10

Definition

Decay constant

If N$N$ is the number of nuclei in the sample and ΔN$\mathrm{\Delta }N$ undergo decay in time Δt$\mathrm{\Delta }t$ then
ΔNΔt∝N${\displaystyle \frac{\mathrm{\Delta }N}{\mathrm{\Delta }t}}\propto N$
or ΔNΔt=λN${\displaystyle \frac{\mathrm{\Delta }N}{\mathrm{\Delta }t}}=\lambda N$
where λ$\lambda$ is called the radioactive decay constant or disintegration constant.

Definition

Relationship between different units of radioactivity

1 becquerel = 1Bq = 1 radioactive decay per second
1 curie = 1 Ci = 3.7 10^{10} Bq (decays per second)

Diagram

Line of stability in a plot of number of neutrons vs protons

Definition

Half life of a radioactive substance

The half-life T1/2$T_{1/2}$ of a radio nuclide is the time at which both N and R have been reduced to one-half their initial values.
Putting R=(1/2)R0$R=(1/2)R_0$ and t=T1/2$t=T_{1/2}$ in the law of radioactive decay, we get:
T1/2=ln2λ=0.693λ$T_{1/2}={\displaystyle \frac{ln2}{\lambda }}={\displaystyle \frac{0.693}{\lambda }}$

Example

Exponential decay of radioactive species at which decay rate at a certain time is involved

Example: A count rate meter is used to measure the activity of a given sample. At one instant, the meter shows 4750$4750$ counts per minute. Five minutes later, it shows 2700$2700$ counts per minute. Find (a) the decay constant and  (b) the half-life of the sample. (log101.760=0.2455)$(lo{g}_{10}1.760=0.2455)$

Solution:
We have,
N=N0e−λt$N=N_0{e}^{-\lambda t}$ (i)
The activity of the sample is given by
dNdt=−λNoe−λt=−λN${\displaystyle \frac{dN}{dt}}=-\lambda N_o{e}^{-\lambda t}=-\lambda N$
i.e., activity is proportional to the number of undecayed nuclei.
At t=0,(dNdt)t=0=−λN$t=0,{\left({\displaystyle \frac{dN}{dt}}\right)}_{t=0}=-\lambda N$
At t=5min,(dNdt)t=5min=−λN$t=5min,{\left({\displaystyle \frac{dN}{dt}}\right)}_{t=5min}=-\lambda N$
∴N0N=(dNdt)t=0(dNdt)t=5min=47502700=1.760$\therefore {\displaystyle \frac{N_0}{N}}={\displaystyle \frac{{\left({\displaystyle \frac{dN}{dt}}\right)}_{t=0}}{{\left({\displaystyle \frac{dN}{dt}}\right)}_{t=5min}}}={\displaystyle \frac{4750}{2700}}=1.760$
a. From Eq. (i), we have
NN0=e−λt${\displaystyle \frac{N}{N_0}}=e^{-\lambda t}$
or N0N=eλt${\displaystyle \frac{N_0}{N}}=e^{\lambda t}$ or logeN0N=λ$lo{g}_e{\displaystyle \frac{N_0}{N}}=\lambda$
⇒λ=1tlogeN0N=2.3026t×log10(N0N)$\Rightarrow \lambda ={\displaystyle \frac{1}{t}}lo{g}_e\frac{N_0}{N}={\displaystyle \frac{2.3026}{t}}\times lo{g}_{10}\left({\displaystyle \frac{N_0}{N}}\right)$ (iii)
Substituting the given values, we have
λ=2.30265×0.2455=0.113min−1$\lambda ={\displaystyle \frac{2.3026}{5}}\times 0.2455=0.113mi{n}^{-1}$
b. Half0life of sample, T=0.6931λ=0.69310.113=6.1min−1$T={\displaystyle \frac{0.6931}{\lambda }}={\displaystyle \frac{0.6931}{0.113}}=6.1mi{n}^{-1}$

Definition

Activity of a radioactive substance

The total decay rate R of a sample of one or more radionuclides is called the activity of that sample.
R=−dNdt=λNe−λt=R0e−λt=λN$R=-{\displaystyle \frac{dN}{dt}}=\lambda N{e}^{-\lambda t}=R_0{e}^{-\lambda t}=\lambda N$
Here R0$R_0$ is the radioactive decay rate at time t = 0, and R is the rate at any subsequent time t.

Definition

Mean value of a radioactive substance

Mean life (τ$\tau$) = sum of life of all atom / total no of atoms present
τ=1λ$\tau ={\displaystyle \frac{1}{\lambda }}$
where λ$\lambda$ is the decay constant.

Example

Fraction of substance decayed in a period of time

Tritium has a half-life of 12.5 y undergoing beta decay.What fraction of a sample of pure tritium will remain undecayed after 25 y.
Solution
By definition of half-life, half of the initial sample will remain undecayed after 12.5 y. In the next 12.5 y, one-half of these nuclei would have decayed. Hence, one fourth of the sample of the initial pure tritium will remain undecayed.

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