Question

Rutherford and Sodi rules in relation to radioactive decay and formula with the help of Rutherford and Sodi rules
Generate [N = Noe ^λt ] ?

Answer 1

Law of Radioactive Decay

When a radioactive material undergoes α, β or γ-decay, the number of nuclei undergoing the decay, per unit time, is proportional to the total number of nuclei in the sample material. So,

If N = total number of nuclei in the sample and ΔN = number of nuclei that undergo decay in time Δt then,

ΔN/ Δt ∝ N

Or, ΔN/ Δt = λN … (1)

where λ = radioactive decay constant or disintegration constant. Now, the change in the number of nuclei in the sample is, dN = – ΔN in time Δt. Hence, the rate of change of N (in the limit Δt→ 0) is,

dN/dt = – λN

Or, dN/N = – λdt

Now, integrating both sides of the above equation, we get,

NN0∫ dN/N = λ tt0∫ dt … (2)

Or, ln N – ln N0 = – λ (t – t0) … (3)

Where, N0 is the number of radioactive nuclei in the sample at some arbitrary time t0and N is the number of radioactive nuclei at any subsequent time t. Next, we set t0 = 0 and rearrange the above equation (3) to get,

ln (N/N0) = – λt

Or, N(t) = N0e– λt … (4)

Equation (4) is the Law of Radioactive Decay.

The Decay Rate

In radioactivity calculations, we are more interested in the decay rate R ( = – dN/dt) than in N itself. This rate gives us the number of nuclei decaying per unit time. Even if we don’t know the number of nuclei in the sample, by simply measuring the number of emissions of α, β or γ particles in 10 or 20 seconds, we can calculate the decay rate. Let’s say that we consider a time interval dt and get a decay count ΔN (= –dN). The Decay rate is now defined as,

R = – dN/dt

Differentiating equation (4) on both sides, we get,

R = λN0 e−λt

Or, R = R0e−λt … (5)

Where, R0 is the radioactive decay rate at time t = 0, and R is the rate at any subsequent time t. Equation (5) is the alternative form of the Law of Radioactive Decay. Now we can rewrite equation (1) as follows,

R = λN … (6)

where R and the number of radioactive nuclei that have not yet undergone decay must be evaluated at the same instant

Half-Life and Mean Life

The total decay rate of a sample is also known as the activity of the sample. The SI unit for measurement of activity is ‘becquerel’ and is defined as,

1 becquerel = 1Bq = 1 decay per second

An older unit, the curie, is still in common use:

1 curie = 1 Ci = 3.7 × 1010 Bq (decays per second)

There are two ways to measure the time for which a radionuclide can last.

Half-life T1/2 – the time at which both R and N are reduced to half of their initial values

Mean life τ – the time at which both R and N have been reduced to, e-1 of their initial values

Answer 2

$\bold{\underline{Rutherford\:and\:Sodi\:rule}}$

The rate of dissolution of any radioactive substance at any given moment
(Rate of decay) at that moment is proportional to the remaining amount of radioactive substance that is the rule of Rutherford Sodi .

Let, The number of atoms NO initially considered (t = 0)

at $\bold{ t}$ time remaining atom = N

$\bold{From\: the\: law \:of\: Rutherford \:and\: sodi -}$

$\frac{dN}{dt} \propto \: N$

$\bold{\huge{\underline{Note}}}$☞The rate at which disruption occurs in a moment is the proportion of the remaining atom after $\bold{t}$ time .

$\bold{\huge{\underline{Formula-}}}$

$\boxed{ \boxed{ \frac{dN}{dt} = - \lambda \: N}}$

$\bold{Origin\: of \:formula\:N=Noe^{-\lambda \:t }\: with \:the \:help \:of\: Rutherford,}$
$\bold{Soddy's\: law}$

$\bold{From\: the\: law \:of \:Rutherford\: and \:Sodi,}$

$\frac{dN}{dt} = -\lambda \: N$

$\frac{dN}{dt}= -\lambda\:d\:t$

$\bold{\underline{On\: the \:consolidation \:of \:both\: sides -}}$

$\displaystyle \int \frac{1}{N} \:dn=\int-\lambda dt + c \\ \\ \implies log_{e}{N}=-\lambda \int dt \: + c \\ \\ \implies log_{e}{ N} = -\lambda \: t \: +c.................1$

Initially - when t = 0 then the remaining quantity N = No, from this value equation (1)

$log_{e}(NO) = - \lambda (0) + c$

$c = log_{e}No$

$\bold{\underline{From \:the \:equation\: first \:}}$

$log_{e}N \: = - \lambda \: t + log_{e}No \: \\ \\ log_{e}N - log_{e}No = - \lambda \: t \\ \\ e {}^{ - \lambda \:t} = ( \frac{N}{No} ) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{N = Noe {}^{ - \lambda \: t} }}$

$\bold{\huge{\underline{Proved}}}$