## Atomic Nucleus Exponential Law Of Radioactive Decay

The statistical law of probability is utilised in the analysis of radioactivity, which is characterised by its spontaneous and stochastic nature. For the radioactive sample, it is impossible to determine which nucleus will be affected.

will disintegrate first, the norcanthesequence of occurrence be ascertained beforehand. Only we can say, that the time rate of disintegration will be directly proportional to the number of radioactive particles present in the sample at that time. Let at time t, the number of radioactive particles present in the sample be N, and in time dt, dN number of particles disintegrate. So, the rate of disintegration is \(\frac{d N}{d t}\) and

⇒ \(\frac{d N}{d t} \propto N \text { or, } \frac{d N}{d t}=-\lambda N\) …………………. (1)

Where A in equation (1) is called the decay constant or radioactive disintegration constant. A is the characteristic ofthe radioactive element used. The negative sign indicates a decrease in several radioactive elements with time.

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**Radioactive decay curve:**

If at the beginning of the count for disintegration that is at t = 0, the number of radioactive particles is N_{2} and after a time interval t, the number of radioactive particles be N, then from equation (1)

⇒ \(\int_{N_0}^N \cdot \frac{d N}{N}=-\int_0^t \lambda d t \text { or, }\left[\log _e N\right]_{N_0}^N=-\lambda t\)

Or, \(\log _e \frac{N}{N_0}=-\lambda t \text { or, } N=N_0 e^{-\lambda t}\) …………………. (2)

The equation N = \(N_0 e^{-\lambda t}\) is the exponential law of radioactive decay. Represents the law graphically. The graph shows that the value of N decreases exponentially with time.

**Decay constant**

**Definition:** The decay constant is the reciprocal of time j during which the number of atoms of a radioactive substance j decreases to – (or 36.8%) ofthe number present initially

Substituting t = 1/λ inequation(2),weget,

N = \(N_0 e^{-\lambda \cdot \frac{1}{\lambda}}=N_0 e^{-1}=\frac{N_0}{e}=0.368 \times N_0\)

**Haff-life**

**Definition:**

The period after which the number of radioactive atoms present in a radioactive sample becomes half of the initial number due to disintegration is called the half-life of that radioactive element.

Like decay constant λ, half-life is also a characteristic of that radioactive element. The half-life of different elements is given below.

**1. Relation between half-life and decay constant:**

Let at the beginning of the count for disintegration i.e., at t = 0 number of radioactive atoms present in a radioactive sample = N_{0}. After a time t this number = N_{0}.

Then according to the exponential law, N = N_{0}e^{–}^{λT }[λ= decay constant]

Now if the half-life of that element = T, then after time T the number of atoms present in the sample

N = \(\frac{N_0}{2}\)

∴ \(\frac{N_0}{2}=N_0 e^{-\lambda T}\)

Or, \(\frac{1}{2}=e^{-\lambda T}\)

e^{λT }= 2

Or, λT = log²_{e}

T = \(\frac{\log _e 2}{\lambda}=\frac{2.303 \log _{10} 2}{\lambda}=\frac{0.693}{\lambda}\) ……………… (3)

Equation (3) gives the relation between the half-life period of the radioactive element and its decay constant. The equation also shows that the half-life period is inversely proportional to the decay constant. Unit of A is per second or s^{-1}

Also, from the relation N = N_{0 }e^{λT}, we get

⇒ \(\frac{N_0}{N}=e^{\lambda t}=\left(e^{\lambda T}\right)^{\frac{t}{T}}=2^{\frac{t}{T}} \text { or, } N=\frac{N_0}{2^{\frac{t}{T}}}\) …………….. (4)

This equation enables one to calculate the number of radioactive particles present after any time interval t,

**2. Significance of half-life:**

Any radioactive substance has a half-life of T, then after time T, 2T,3T, the fraction of the initial amount (N_{0}) that disintegrates and the fraction that remains.

The table clearly shows that no radioactive substance can completely disintegrate and so there is no complete life of such a substance. To express the radioactive properties, therefore, we need to know the mean life ofthe radioactive substance.

**Mean life or average life**

The mean life or average life of a radioactive element is defined as the ratio ofthe total lifetime of all the radioactive atoms to the total number of such atoms in it

Let us consider a radioactive element containing N_{0} number of atoms at time t = 0. Let the number of atoms left at time t be N. Suppose a small number of atoms, dN disintegrates further in a small time dt. Therefore, the lifetime of each of these dN atoms lies between t and (t+dt). Since it is small, we can say. dN atoms lived for a time of t.

So total lifetime of dN atoms = tdN

Total lifetime of all atoms = \(\int_0^{N_0} t d N\)

Thus the mean life or average life of a radioactive element is the reciprocal of the radioactive constant.

**Relation between half-life and mean life:**

The mean life of a radioactive element is the reciprocal of the decay constant i.e., mean life, τ = 1/λ Hence from equation (3),

Half-life, T = 0.693r or, τ = 1.443T ………………… (5)

Equation (5) gives us the relation between half-life (T) and mean life (τ). The characteristics of radioactive elements can be represented by mean life instead of half-life in some cases. Ra-226 has a half-life T = 1600 y. Hence, its mean life is (1600 × 1.443) y or about 2300 y.

## Atomic Nucleus Exponential Law Of Radioactive Decay Numerical Examples

**Example 1. The half-life of a radioactive substance is 1 y. After n2 y, what will the amount of the substance that will be disintegrated?
**

**Solution:**

After 1 y, the remaining substance| = \(\frac{1}{2}\) part

∴ After 2y, the amount of substance that will remain

= \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}\) part

Amount of the substance that is disintegrated after 2y

= \(1-\frac{1}{4}=\frac{3}{4}\) part

**Example 2. In 8000 y a radioactive substance reduces to \(\frac{1}{32}\)th part. Determine its half-life. **

**Solution:**

Let the initial amount of radioactive substance be 1 and the half-life is T

According to the question

5 T = 8000

Or, T = \(\frac{8000}{5}\)

= 1600 y

**Example 3. A radioactive material reduces to \(\frac{1}{8}\)th of its Initialr-rrajÿÿtt in 18000 y. Find its half-life period**

**Solution:**

Here, t = 18000 y and \(\frac{N}{N_0}=\frac{1}{8}=\frac{1}{2^3}\)

From equation N = \(\frac{N_0}{2^{t / T}}\) , we get

= \(\frac{1}{2^3}=\frac{1}{2^{18000 / T}}\)

Or, 3T = 18000 or, T = 6000 y

**Alternative method:**

Let the initial amount of radioactive substance be 1 and the half-life is T.

∴ According to question

3T = 18000 Or, T = \(\frac{18000}{3}\)

= 6000 y

**Example 4. An accident In the laboratory deposits some amount of radioactive material of half-life 20d on the floor and the walls. Testing reveals that the level of radiation Is 32 times the maximum permissible level. After how many days will it be safe to use the room**

**Solution: **

Half-life, T= 20d

The number of days after which the room can be used safely

= 5T = 5 × 20 = 100 d

**Example 5. The half-life of thorium is 1.5 × 10 ^{10} y. How much time is needed for 20% of thorium to disintegrate?**

**Solution:**

Let the initial mass of thorium = N_{0}

If in time t 20% of the thorium is disintegrated then, the amount of thorium that disintegrates

= \(N_0 \times \frac{20}{100}=0.2 N_0\)

Amount of thorium left

N = N_{0}– 2N_{0} = 0.8 N_{0}

N = N_{0}e^{– λt}

Now N = \(e^{\lambda t}=\frac{N_0}{N}=\frac{N_0}{0.8 N_0}\)

= 1.25

λt = \(\lambda t=\log _e(1.25)\)

= 0. 223

Or, \(\frac{0.693}{T} \cdot t\)

= 0.223

Since \(T \text { (half-life) }=\frac{0.693}{\lambda}\)

Or, \(\frac{T}{0.693} \times 0.223\)

= Or, \(\frac{1.5 \times 10^{10} \times 223}{693}\)

= 0.48 × 10^{10} y (approx)

**Alternative method:**

N = 0.8 N

Also N = \(\frac{N_0}{2^{t / T}}\)

Or, 0.8 \(=\frac{1}{2^{t / T}} \text { or, } 2^{t / T}=5 / 4\)

t/T = \(\log _2 5 / 4=\frac{\log _{10} 5 / 4}{\log _{10} 2}\)

= \(\frac{0.0969}{0.3010}\)

= 0.322

Or, t = 0.322T

= 0.322 × 1.5 × 10^{10} y

= 0.48 × 10^{10} y (approx).

**Example 6. The half-life of radium is 1500 y. In how many years will 1 g of pure radium reduce by 1 mg **

**Solution:**

Let the time in which lg radium will reduce by 1 mg = t

So, remaining mass of radium = 1- 0.001 = 0.999 g

Now, assuming the initial mass is N_{0}, and in time t mass becomes N then

N/ N_{0 }= 0.999 /1 = 0.999

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

Or, = \(e^{\lambda t}=\frac{N_0}{N}=\frac{1}{0.999}\) = 1.001(approx)

λt = \(\lambda t=\log _e(1.001)\) = 0.001 (approx)

Or, = \(\frac{0.693}{T} \cdot t\)

= 0.0001

Since = \(\text { half-life, } T=\frac{0.693}{\lambda}\)

Or, t = \(\frac{T}{0.693} \times 0.001\)

= \(\frac{1500}{693}\)

= 2.16y (approx)

**Example 7. State the law of radioactive decay. Three fourth of a radioactive sample decays in ¾ s. What is the half-life of the sample?**

**Solution:**

The rate of decay of a radioactive sample concerning time is proportional to the number of radioactive atoms present in the sample at that instant. This is the law of radioactive decay. As per this law, if N_{0} is the number of atoms of a certain radioactive element initially, and N is its number after a time t, then

N = N_{0 }e^{– λt} (where λ = radioactive decay constant)

Given \(\frac{3}{4}\) of the simple decay in \(\frac{3}{4}\) s

So, 2T = \(\frac{3}{4}\) s Or, T = \(\frac{3}{8}\)s

**Example 8. A radioactive isotope X with a half-life of 1.5 × 10 ^{9} y decays into a stable nucleus Y. A rock sample contains both elements X and Y in a ratio of 1:15. Find the age of the rock**

**Solution:**

X → Y (stable)

Let the quantity of X and Y in the sample be N_{x }and N_{ y }respectively.

⇒ \(\frac{N_x}{N_y}=\frac{1}{15}\) Or, \(\frac{N_x}{N_x+N_y}=\frac{1}{16}\)

Or, \(\frac{N}{N_0}=\frac{1}{16}\)

(\(V_0=N_x+N_y \text { and } N_x=N\))

We know that, N = \(\left[N_0=N_x+N_y \text { and } N_x=N\right]\)

∴ \(e^{\lambda t}=\frac{N_0}{N}\) = 16

Or, t = \(\frac{4 \ln 2}{\lambda}=\frac{4 \ln 2 \times t_{1 / 2}}{\ln 2}\)

or, \(\lambda=\frac{\ln 2}{t_{1 / 2}}\)

Or, t = \(4 \times 1.5 \times 10^9 y=6 \times 10^9 y\)

Age of the rock = 6 × 10^{9} y .