## What is the Radioactive Decay Formula?

A parent nucleus splits into two or more daughter nuclei to reach the stage of stability. So the unstable nucleus is considered radioactive and while splitting, it loses energy in the form of radiation.

So the whole process initiating from subdivision to loss of energy is the radioactive decay.

The three most common types of decays are:

• Alpha decay

• Beta-decay (β - decay)

• Gamma decay (𝛾 - decay)

There are certain radioactive equations for three of these, say, for the gamma decay process, we have the gamma decay formula, proceeding with this, we have the activity of radioactive substance formula.

Also, the formula for half-life decay helps us determine the time needed for half of the original population of radioactive atoms to decay, which we will understand with the help of the radioactive half-life formula.

As per the activity of radioactive substance formula, the average number of radioactive decays per unit time or the change in the number of radioactive nuclei present is given as:

A   =  -  dN/dt

Here,

• A is the total activity

• N is the number of particles

• T = time taken for the whole activity to complete

Also, we understand the following key points from the above radioactivity equation:

• The total activity relies entirely on the number of nuclei present, as we can see A dN....(a)

• During the radioactive decay, A decreases with time, as we can see that A 1/dt…..(b)

Below is the graph representing (a) and (b) definitions:

Now, let us have a look at alpha, beta, and gamma radioactive equations.

In the years 1899 and 1900, a British Ernest Rutherford (working at McGill University in Montreal, Canada) and the French Physicist named Paul Villard (working in Paris) did experimental investigations on electromagnetic radiation and separated them into three kinds.

Further, Rutherford named them alpha, beta, and gamma rays depending on the penetration of matter and deflection by a magnetic field.

• Alpha decay formula

• Beta decay formula

• Gamma decay formula

### Alpha Decay Formula

Alpha decay results in the emission of α-particles from the radioactive nucleus.

For example, the alpha decay of  92U238 into   90Th234  is as follows:

 92U238        →    α-decay    ⇾  90Th234    +   2He 4…(1)

In the above radioactive decay formula, we notice the following things:

1. The Uranium nucleus emits an α-particle, and therefore, its mass and charge reduced, shown in the following equation:

Mass number:  238 -  4 = 234

Charge number: 92 - 2 = 90

Following this, a new element formed is Thorium (Th).

In general, the radioactivity equation (1) can be represented as:

zXA   ⇾   z - ₂YA-2  +  ₂He⁴ + Q

Also, we see that after a spontaneous α-decay process, the total mass of  90Th234 and  2He 4 was less than 92U238.

This means, the total mass-energy (Q) also decreases, equivalent to the difference between the Initial mass-energy and the final mass-energy, stated as:

Q = (mx - m - mHe)

Q = (mx - my - mHe). c2

In this equation, Q is the disintegration energy, which is shared by the daughter nucleus ‘y’ and an alpha particle He.

So, the alpha particle emission occurs in the following manner:

### Beta Decay Formula

Beta decay of Thorium, i.e.,  90Th234 emits a β-particle, where the mass number of the

daughter nucleus remains invariant, while the charge increments by 1, therefore, a new element Palladium 91Pa234  forms in the following way:

90Th234   91Pa234 + -1e0 (β-particle)

Here,

2. The mass number of Palladium is:

234 - 0 = 234

Its charge number becomes 91 (90+1).

The general radioactivity equation for the beta decay process is:

Z XA   z+1YA + -1e0 + Q

Q = the energy released in β-decay.

### Gamma Decay Formula

We know that gamma rays are emitted during the decay of radioactive atomic nuclei and definite subatomic particles.

These powerful rays are produced by the hottest and most energetic objects in the universe and are present in the electromagnetic spectrum.

Below is the gamma decay equation of Technetium-99m to Technetium-99:

43Tc99m      →      43Tc99      +   0γ0   (Gamma radiation) ….(2)

Another example that initiates from the β-decay of 27Co60 turning into an exciting 28Ni60 nucleus, the radioactive decay equation for the same with the energy released is as follows:

27Co60   28Ni60**   + -1e0……(3)

So, when this exciting nucleus reaches the ground state, as a result, gamma rays are emitted with a release of energy in Mega electron Volts.

28Ni60** 28Ni60* + Eγ ( = 1.17 MeV)……(4)

28Ni60* 28Ni + Eγ (= 1.33 MeV)……(5)

The formula for half-life decay is:

$N(t)=N_{0}(\frac{1}{2})^{\frac{t}{t_{1/2}}}$……(6)

Here,

• N (t) is a function of time, which shows the amount of substance remaining after the decay in a given time.

• N is the initial quantity of the substance

• t  is the time elapsed, and

• t1/2  is the half-life of the decaying component

### Definition of the Half-Life:

When half of the radioactive atom undergoes the decay process, the time needed for a quantity to reduce to half of its initial value is the half-life. When talking about the decay of half of the radioactive atoms, the time taken is the radioactive half-life.

For this, we have a radioactive half-life formula:

$t_{1/2}=\frac{0.693}{\lambda }$

Here, λ is the decay constant.

Now, let us understand the decay constant formula:

Let’s suppose that ‘N’ is the size of a population of radioactive atoms at a given time ‘t,’ and dN is the amount by which the population of the radioactive atom decreases in time dt; therefore,  the rate of change is given by the following equation:

dN/dt     = - λ N,  (λ  = decay constant)

 Number of Radioactive Half-Lives Elapsed (Passed) Remaining Fraction % Age Remaining 0 1/1 100% 1 1/2 50% 2 1/4 25% 3 1/8 12.5% 4 1/16 6.25% 5 1/32 3.125% 6 1/64 1.5625% 7 1/128 0.78125% 8 1/256 0.390625% .... n 1/2n 100/2n

The graph of the above table is as follows:

Let’s describe equation (6) in an exponential form:

$N(t)=N_{0}(\frac{1}{2})^{\frac{t}{t_{1/2}}}$

$N(t)=N_{0}e^{\frac{-t}{\tau}}$

Here,

τ =  A mean lifetime of the decaying quantity, which is positive

λ is also positive

The three parameters (t1/2, τ , and λ are all directly related to each other in the following way:

$t_{1/2}=\frac{ln(2)}{\lambda }$  =   (ln (2). τ

We know that the value of the natural logarithmic of 2 is 0.693, so rewriting equation (7):

$t_{1/2}=\frac{0.693}{\lambda }$

### Conclusion

A phenomenon in which a ‌heavy‌ ‌unstable element‌ ‌disintegrates‌ ‌itself‌ into two or more daughter nuclei ‌without‌ ‌being‌ forced‌ ‌by‌ ‌any‌ ‌external‌ ‌agent‌ ‌to‌ ‌do‌ ‌so.‌ The process involves the emission of the following particles:

• α particle

• β particle

• 𝛾 particle

## Solved Examples

1. A radioactive sample has 6.0 x 1018 active nuclei at a certain instant. How many of these nuclei will still be in the same active state after two half-lives?

Sol.

In one half-life, the number of active nuclei reduces to half the original number. Thus, in two half-lives the number is reduced to $\lgroup \dfrac{1}{2}\rgroup \times \lgroup \dfrac{1}{2}\rgroup$ of the original number. The number of remaining active nuclei is, therefore,

$N_{active} = 6.0 \times 10^{18} \times \lgroup \dfrac{1}{2}\rgroup \times \lgroup \dfrac{1}{2}\rgroup$

$N_{active} = 1.5 \times 10^{18}$

1. The mean lives of an unstable nucleus in two different decay processes are 1620 yr and 405 yr, respectively. Find out the time during which three-fourth of a sample will decay.

Sol:

Given,

Mean life of Process- 1 (t1) = 1620 yr,

Mean life of Process- 2 (t2) = 405 yr,

Let decay constants of process-1 and process- 2 be λ1 and λ2, respectively. Then,

$\lambda_1 = \dfrac{1}{t_1}$

$\lambda_2 = \dfrac{1}{t_2}$

If the effective decay constant is λ, then

λN = λ1N + λ2

λ = λ1 + λ2

$\lambda = \dfrac{1}{t_1}+\dfrac{1}{t_2}$

$\lambda = \dfrac{1}{1620}+\dfrac{1}{405}$ year-1

$\lambda = \dfrac{1}{324}$ year-1

Now, when three- fourth of the sample will decay. The sample remaining will be one- fourth of the total sample. If the initial sample was N0, then final sample will have N0/4,

So, $\dfrac{N_0}{4} = N_0e^{-\lambda t}$

Applying logarithms on both the sides, we get,

$-\lambda t = \ln\lgroup\dfrac{1}{4}\rgroup$ = -1.386

t = 449 yr

Therefore, the time during which three-fourth of a sample will decay will be 449 yr.

## FAQs on Radioactive Decay Formula

1. What is the Significance of a Half-Life?

The term ‘half-life’ is mostly used in nuclear physics to express how quickly unstable atoms undergo radioactive decay to stable nuclei, also, how long stable atoms survive.

Also, half-life specifically characterizes the type of exponential or non-exponential decay.

For instance, the medical sciences are more inclined to the biological half-life of drugs and other chemicals in the human body.

2. How are Gamma Rays Produced on the Earth?

Naturally, gamma rays are produced on the earth by the emissions produced by various objects, such as

Nuclear explosions, lightning, radioactive decay activity

Nuclear reactions such as fusion, fission, alpha decay, and 𝛾 decay.

3. What is the Most Dangerous Radiation?

Among the three radiation types, alpha radiation is the most dangerous because it gets easily absorbed by cells. However, beta and gamma radiation are not that dangerous because they are less absorbed by cells and can easily pass through them.

4. What is the Use of Gamma Rays?

Gamma rays travel at a speed of 3 x 108 m/s, their uses are:

Cancer treatment

Tracking fluid flows

Pasteurization

Food preservation

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