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Radioactivity and Decay Laws



Radioactivity

Radioactivity is the spontaneous emission of radiation from unstable atomic nuclei. Certain isotopes of elements have nuclei that are not stable. These unstable nuclei undergo spontaneous transformations, emitting particles and/or electromagnetic radiation to attain a more stable configuration. This process is called radioactive decay. The substances that exhibit radioactivity are called radioactive substances.

Radioactivity was discovered by Henri Becquerel in 1896 while studying phosphorescence in uranium salts. Later, Pierre and Marie Curie did pioneering work in isolating new radioactive elements like Polonium and Radium and characterised the nature of the emitted radiations.

The radiations emitted by radioactive substances were found to be primarily of three types, which were named Alpha ($\alpha$), Beta ($\beta$), and Gamma ($\gamma$) rays based on their penetrating power and deflection in electric and magnetic fields.


Law Of Radioactive Decay ($ N(t) = N_0 e^{-\lambda t} $)

Radioactive decay is a random process at the level of individual nuclei. We cannot predict when a particular nucleus will decay. However, for a large sample of radioactive nuclei, the decay process follows a definite statistical law.

The Law of Radioactive Decay states that:

"The rate of disintegration of a radioactive substance at any instant is directly proportional to the number of radioactive atoms present in the sample at that instant."

Let $N$ be the number of radioactive nuclei present in a sample at time $t$. The rate of disintegration is $-\frac{dN}{dt}$ (the negative sign indicates that $N$ is decreasing). According to the law:

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

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

Where $\lambda$ is a constant called the decay constant or disintegration constant. The decay constant is characteristic of the particular radioactive isotope and represents the probability of decay per unit time for a single nucleus. A larger $\lambda$ means a faster decay rate.

The differential equation can be written as $\frac{dN}{N} = -\lambda dt$. Integrating both sides:

$ \int_{N_0}^{N(t)} \frac{dN}{N} = \int_0^t -\lambda dt $

Where $N_0$ is the number of radioactive nuclei at time $t=0$, and $N(t)$ is the number of radioactive nuclei remaining at time $t$.

$ [\ln N]_{N_0}^{N(t)} = -\lambda [t]_0^t $

$ \ln N(t) - \ln N_0 = -\lambda t $

$ \ln \left(\frac{N(t)}{N_0}\right) = -\lambda t $

Taking the exponential of both sides:

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

$ N(t) = N_0 e^{-\lambda t} $

This is the Radioactive Decay Law. It describes how the number of radioactive nuclei in a sample decreases exponentially with time. The number of remaining nuclei never truly reaches zero in finite time according to this model, although the rate of decay becomes very small.

The rate of decay ($dN/dt$) is also called the activity ($R$) of the sample (discussed in Section I3).


Alpha Decay ($\alpha$-Decay)

In alpha decay, an unstable nucleus emits an alpha particle ($\alpha$). An alpha particle is a helium nucleus ($_2^4He$), consisting of 2 protons and 2 neutrons.

When a parent nucleus $_Z^A X$ undergoes alpha decay, it transforms into a daughter nucleus $_ {Z-2}^{A-4} Y$ and emits an alpha particle.

$ _Z^A X \longrightarrow _{Z-2}^{A-4} Y + _2^4He $

Properties of Alpha Particles:

Alpha decay typically occurs in heavy, unstable nuclei. The energy released in alpha decay ($Q$-value) appears as kinetic energy of the daughter nucleus and the alpha particle. The daughter nucleus is usually in an excited state and may subsequently emit gamma rays.


Beta Decay ($\beta$-Decay)

In beta decay, a neutron in the nucleus is converted into a proton, or a proton is converted into a neutron. There are three main types of beta decay: $\beta^-$, $\beta^+$, and electron capture.

1. Beta-minus decay ($\beta^-$): A neutron is converted into a proton, an electron (called a beta particle, $\beta^-$), and an electron antineutrino ($\bar{\nu}_e$).

$ n \longrightarrow p + e^- + \bar{\nu}_e $

The parent nucleus $_Z^A X$ transforms into a daughter nucleus $_{Z+1}^A Y$ with the same mass number but atomic number increased by 1.

$ _Z^A X \longrightarrow _{Z+1}^A Y + e^- + \bar{\nu}_e $

$\beta^-$-decay occurs in neutron-rich nuclei. The electron and antineutrino are created at the moment of decay; they do not exist in the nucleus beforehand.

2. Beta-plus decay ($\beta^+$): A proton is converted into a neutron, a positron (a positively charged electron, $\beta^+$), and an electron neutrino ($\nu_e$).

$ p \longrightarrow n + e^+ + \nu_e $

The parent nucleus $_Z^A X$ transforms into a daughter nucleus $_{Z-1}^A Y$ with the same mass number but atomic number decreased by 1.

$ _Z^A X \longrightarrow _{Z-1}^A Y + e^+ + \nu_e $

$\beta^+$-decay occurs in proton-rich nuclei. The positron and neutrino are created at the moment of decay.

3. Electron Capture: The nucleus captures an inner atomic electron. This electron combines with a proton in the nucleus to form a neutron and an electron neutrino.

$ p + e^- \longrightarrow n + \nu_e $

The parent nucleus $_Z^A X$ transforms into a daughter nucleus $_{Z-1}^A Y$. This is an alternative to $\beta^+$-decay for proton-rich nuclei.

Properties of Beta Particles ($\beta^-$ and $\beta^+$):


Gamma Decay ($\gamma$-Decay)

In gamma decay, a nucleus in an excited energy state releases energy in the form of a gamma ray photon ($\gamma$). The nucleus does not change its composition (number of protons or neutrons); it simply transitions from a higher energy state to a lower energy state.

$ _Z^A X^* \longrightarrow _Z^A X + \gamma $ (where $X^*$ denotes an excited nucleus)

Gamma decay often follows alpha or beta decay, as the daughter nucleus is frequently produced in an excited state.

Properties of Gamma Rays:


Example 1. Write the nuclear equation for the alpha decay of Uranium-238 ($_{92}^{238}U$). Identify the daughter nucleus.

Answer:

Given:

Parent nucleus is Uranium-238, $_{92}^{238}U$. Here $Z=92$, $A=238$.

Alpha decay involves the emission of an alpha particle, $_2^4He$.

Let the daughter nucleus be $_ {Z'}^{A'} Y$. The nuclear equation for alpha decay is $_Z^A X \longrightarrow _{Z-2}^{A-4} Y + _2^4He$.

Applying this to Uranium-238:

Atomic number of daughter nucleus, $Z' = Z - 2 = 92 - 2 = 90$.

Mass number of daughter nucleus, $A' = A - 4 = 238 - 4 = 234$.

The element with atomic number 90 is Thorium (Th).

So, the daughter nucleus is Thorium-234, $_{90}^{234}Th$.

The nuclear equation is:

$ _{92}^{238}U \longrightarrow _{90}^{234}Th + _2^4He $

The daughter nucleus is Thorium-234 ($_{90}^{234}Th$).



Half-life ($ T_{1/2} = \ln(2)/\lambda $) and Mean life ($ \tau = 1/\lambda $)

The radioactive decay law $N(t) = N_0 e^{-\lambda t}$ describes how the number of radioactive nuclei decreases over time. While the number of nuclei never reaches zero, it is useful to define characteristic times related to the decay rate, such as half-life and mean life.


Half-life ($T_{1/2}$)

The half-life ($T_{1/2}$) of a radioactive substance is the time required for the number of radioactive nuclei in a sample to reduce to half of its initial value.

At time $t = T_{1/2}$, the number of remaining nuclei $N(T_{1/2}) = N_0 / 2$.

Using the decay law $N(t) = N_0 e^{-\lambda t}$:

$ \frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} $

Dividing by $N_0$:

$ \frac{1}{2} = e^{-\lambda T_{1/2}} $

Taking the natural logarithm of both sides:

$ \ln\left(\frac{1}{2}\right) = -\lambda T_{1/2} $

Since $\ln(1/2) = \ln(1) - \ln(2) = 0 - \ln(2) = -\ln(2)$:

$ -\ln(2) = -\lambda T_{1/2} $

$ \ln(2) = \lambda T_{1/2} $

Solving for the half-life $T_{1/2}$:

$ T_{1/2} = \frac{\ln(2)}{\lambda} $

Since $\ln(2) \approx 0.693$,

$ T_{1/2} \approx \frac{0.693}{\lambda} $

The half-life is inversely proportional to the decay constant $\lambda$. A larger $\lambda$ (faster decay) corresponds to a shorter half-life. Half-lives of radioactive isotopes vary widely, from fractions of a second to billions of years.


Mean Life ($\tau$)

While half-life is the time for half the nuclei to decay, the mean life ($\tau$) or average lifetime of a radioactive nucleus is the average time that a radioactive nucleus exists before it decays.

The mean life is related to the decay constant by:

$ \tau = \frac{1}{\lambda} $

The derivation of this formula involves integral calculus, averaging over the lifetimes of all nuclei in a large sample.


Relation between Half-life and Mean Life

From the formulas for half-life and mean life:

$ T_{1/2} = \frac{\ln(2)}{\lambda} $ and $ \tau = \frac{1}{\lambda} $

We can express half-life in terms of mean life:

$ T_{1/2} = \ln(2) \cdot \tau \approx 0.693 \tau $

The half-life is approximately 69.3% of the mean life.

At time $t = \tau$, the number of remaining nuclei is $N(\tau) = N_0 e^{-\lambda \tau} = N_0 e^{-(\frac{1}{\tau})\tau} = N_0 e^{-1} \approx 0.368 N_0$. So, in one mean life, the number of nuclei reduces to about 36.8% of the initial value (whereas in one half-life, it reduces to 50%).


Exponential Decay Steps

Using the concept of half-life, we can describe the decay process in terms of successive half-life periods.


Example 1. The half-life of a radioactive isotope is 5 days. If a sample initially contains 1000 radioactive nuclei, how many nuclei will remain after 15 days?

Answer:

Given:

Half-life, $T_{1/2} = 5 \, days$

Initial number of nuclei, $N_0 = 1000$

Time elapsed, $t = 15 \, days$

Method 1: Using the concept of half-lives.

Number of half-lives elapsed = $n = t / T_{1/2} = 15 \, days / 5 \, days = 3$.

After $n$ half-lives, the number of remaining nuclei is $N = N_0 / 2^n$.

$ N = 1000 / 2^3 = 1000 / 8 = 125 $

After 15 days, 125 nuclei will remain.

Method 2: Using the decay law $N(t) = N_0 e^{-\lambda t}$.

First, find the decay constant $\lambda$ from the half-life: $\lambda = \ln(2) / T_{1/2}$.

$ \lambda = \frac{0.693}{5 \, days} = 0.1386 \, days^{-1} $

Now, substitute this and the time $t=15$ days into the decay law:

$ N(15) = 1000 \, e^{-(0.1386 \, days^{-1}) \times (15 \, days)} $

$ N(15) = 1000 \, e^{-2.079} $

$ e^{-2.079} \approx 0.125 $

$ N(15) = 1000 \times 0.125 = 125 $

After 15 days, 125 nuclei will remain. Both methods give the same result.


Example 2. Calculate the mean life of a radioactive substance whose half-life is 10 hours.

Answer:

Given:

Half-life, $T_{1/2} = 10 \, hours$

We use the relation between half-life and mean life: $T_{1/2} = \ln(2) \cdot \tau$.

Rearranging to find the mean life $\tau = T_{1/2} / \ln(2)$.

$ \tau = \frac{10 \, hours}{\ln(2)} $

$ \tau \approx \frac{10}{0.693} \, hours \approx 14.43 \, hours $

The mean life of the radioactive substance is approximately 14.43 hours.



Activity of a Radioactive Sample ($ R = -\frac{dN}{dt} = \lambda N $)

In practice, when dealing with radioactive substances, we are often interested in the rate at which decays are occurring, rather than the exact number of undecayed nuclei. This rate of decay is called the activity of the sample.


Definition of Activity

The activity ($R$) of a radioactive sample is defined as the rate of disintegration (decay) of the nuclei in the sample per unit time. It is equal to the magnitude of $dN/dt$, where $N$ is the number of radioactive nuclei present at time $t$.

From the decay law, the rate of disintegration is given by:

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

The activity $R$ is the magnitude of this rate:

$ R = \left|-\frac{dN}{dt}\right| = \lambda N $

So, the activity of a radioactive sample at any time is directly proportional to the number of radioactive nuclei present at that time, with the decay constant $\lambda$ as the proportionality constant.

Since the number of nuclei $N$ decreases exponentially with time ($N(t) = N_0 e^{-\lambda t}$), the activity $R$ also decreases exponentially with time:

$ R(t) = \lambda N(t) = \lambda (N_0 e^{-\lambda t}) = (\lambda N_0) e^{-\lambda t} $

The initial activity at $t=0$ is $R_0 = \lambda N_0$.

$ R(t) = R_0 e^{-\lambda t} $

The activity decays with the same decay constant $\lambda$ (and thus the same half-life and mean life) as the number of nuclei.


Units of Activity

The SI unit of activity is the becquerel (Bq), named after Henri Becquerel. One becquerel is defined as one disintegration per second (1 dps).

$ 1 \, Bq = 1 \, \text{disintegration per second (dps)} $

A larger, older unit of activity is the curie (Ci), named after Pierre and Marie Curie. It was originally defined as the activity of 1 gram of Radium-226.

$ 1 \, Ci = 3.7 \times 10^{10} \, \text{disintegrations per second} = 3.7 \times 10^{10} \, Bq $

This is a very large unit. Activities are often expressed in millicuries (mCi) or microcuries ($\mu$Ci).


Relation to Half-life and Mean Life

The activity is directly related to the number of nuclei and the decay constant. Since $\lambda = \ln(2)/T_{1/2} = 1/\tau$, we can also write the activity in terms of half-life or mean life:

$ R = \frac{\ln(2)}{T_{1/2}} N = \frac{0.693}{T_{1/2}} N $

$ R = \frac{N}{\tau} $

These formulas are useful for calculating the activity if the number of nuclei and the half-life or mean life are known.


Example 1. A radioactive sample has a decay constant of $0.01 \, s^{-1}$. If the sample contains $10^{20}$ radioactive nuclei, calculate its activity in Becquerel.

Answer:

Given:

Decay constant, $\lambda = 0.01 \, s^{-1}$

Number of radioactive nuclei, $N = 10^{20}$

The activity ($R$) is given by $R = \lambda N$.

$ R = (0.01 \, s^{-1}) \times (10^{20}) $

$ R = 10^{-2} \times 10^{20} \, s^{-1} = 10^{18} \, s^{-1} $

Since 1 Bq = 1 disintegration per second, the activity is:

$ R = 10^{18} \, Bq $

The activity of the sample is $10^{18}$ Becquerel.


Example 2. The half-life of a radioactive isotope is 8 days. Calculate its activity if a sample contains $2.5 \times 10^{16}$ nuclei.

Answer:

Given:

Half-life, $T_{1/2} = 8 \, days$

Number of nuclei, $N = 2.5 \times 10^{16}$

We need to calculate the activity $R = \lambda N$. First, we need to find the decay constant $\lambda$ in reciprocal seconds (for Bq unit). The half-life needs to be converted to seconds.

$ T_{1/2} = 8 \, days \times 24 \, hours/day \times 60 \, minutes/hour \times 60 \, seconds/minute $

$ T_{1/2} = 8 \times 24 \times 3600 \, s = 691200 \, s = 6.912 \times 10^5 \, s $

Now calculate the decay constant $\lambda = \ln(2) / T_{1/2} \approx 0.693 / T_{1/2}$.

$ \lambda = \frac{0.693}{6.912 \times 10^5 \, s} \approx 0.1002 \times 10^{-5} \, s^{-1} = 1.002 \times 10^{-6} \, s^{-1} $

Now calculate the activity $R = \lambda N$.

$ R = (1.002 \times 10^{-6} \, s^{-1}) \times (2.5 \times 10^{16}) $

$ R = (1.002 \times 2.5) \times 10^{10} \, Bq $

$ R \approx 2.505 \times 10^{10} \, Bq $

The activity of the sample is approximately $2.505 \times 10^{10}$ Becquerel.

We can also express this in Curies. $1 \, Ci = 3.7 \times 10^{10} \, Bq$.

$ R = \frac{2.505 \times 10^{10} \, Bq}{3.7 \times 10^{10} \, Bq/Ci} \approx 0.677 \, Ci $

The activity is approximately 0.677 Curies.