By learning Physics Topics, we can gain a deeper appreciation for the natural world and our place in it.
What is Mean Life of Radioactive Substance?
Statistical law of probability is applied to the study of radioac-tivity as it is a spontaneous but random process. This means that for a radioactive sample, neither can it be predicted which nucleus will disintegrate first, nor can the sequence of occurrence be ascertained before hand. Only we can say, 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
\(\frac{d N}{d t}\) ∝ N or, \(\frac{d N}{d t}\) = -λN ……. (1)
where λ in equation (1) is called decay constant or radioactive disintegration constant. A is the characteristic of the radioactive element used. Negative sign indicates the decrease in number of radioactive element with time.
Radioactive decay curve: If at the beginning of the count for disintegration that is at t = 0, the number of radioactive particles be N0 and after a time interval t, the number of radioactive particles be N, then from equation (1)
The equation N = N0eλt is the exponential law of radioactive decay. Fig. represents the law graphically. The graph shows that the value of N decreases exponentially with time.
Decay constant: Substituting t = \(\frac{1}{\lambda}\) in equation (2), we get,
N = N0\(e^{-\lambda \cdot \frac{1}{\lambda}}\) = N0e-1 = \(\frac{N_0}{e}\) = 0.368 × N0
Definition: The decay constant is the reciprocal of time during which the number of atoms of a radioactive substance decreases to \(\frac{1}{e}\) (or 36.8%) of the number present initially.
Half-life:
Definition: The time period after which the number of radioactive atoms present in a radioactive sample becomes half of its initial number due to disintegration is called 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.
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 = N0. After a time t this number = N.
Then according to the exponential law,
N = N0eλt [λ = decay constant]
Now if half-life of that element = T, then after time T the num-ber of atoms present in the sample N = \(\frac{N_0}{2}\) [Fig],
fraction that remains.
∴ \(\frac{N_0}{2}\) = N0e-λt or, eλT = 2
or, λT = loge2
or, T = \(\frac{\log _e 2}{\lambda}\) = \(\frac{2.303 \log _{10} 2}{\lambda}\) = \(\frac{0.693}{\lambda}\) …… (3)
Equation (3) gives the relation between half-life period of the radioactive element and its decay constant. The equation also shows that, half-life period is inversely proportional to the decay constant. Unit of λ is per second or s-1.
Also, from the relation N = N0e-λt, we get
or, \(\frac{N_0}{N}\) = eλt = \(\left(e^{\lambda T}\right)^{\frac{t}{T}}\) = \(2^{\frac{t}{T}}\) 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.
Significance of half-life: TABLE-5 shows that if any radio active substance has a half-life T, then after time T, 2T, 3T, the fraction of the initial amount (N0) that disintegrates and the fraction that remains.
Table-5
The table clearly shows that no radioactive substance can completely disintegrate and so there is no complete or full life of such a substance. To express the radioactive properties therefore, we need to know the mean life of the radioactive substance.
Mean life or average life: The mean life or average life of a radioactive element is defined as the ratio of the total life dine of all the radioactive atoms to the total number of such atoms in it.
Let us consider a radioactive element containing N0 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 disintegrate further in a small time dt. Therefore, the life time of each of these dN atoms lies between t and (t + dt). Since dt is small, we can say dN atoms lived for a time of t.
So total life time of dN atoms = tdN
Total life time 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: Mean life of a radioactive element is the reciprocal of the decay constant i.e., mean life, \(\tau\) = \(\frac{1}{\lambda}\). Hence from equation (3),
half-life, T = 0.693\(\tau\) or, \(\tau\) = 1.443T ….. (5)
Equation (5) gives us the relation between half-life (T) and mean life (\(\tau\)). The characteristic of radioactive element can be represented by mean life instead of half-life in some cases. Ra-226 has half-life T = 1600 y. Hence, its mean life is (1600 × 1.443)y or about 2300 y.
Numerical Examples
Example 1.
The half-life of a radioactive substance is 1 y. After 2 y, what will the amount of the substance that will be disintegrated? [WBJEE 2000]
Solution:
After 1 y, the remaining substance = \(\frac{1}{2}\)part
∴ After 2 y, the amount of substance that will remain = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\) 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 initial amount of radioactive substance be 1 and half-life is T.
∴ According to question,
5T = 8000 or, T = \(\frac{8000}{5}\) = 1600 y
Example 3.
A radioactive material reduces to \(\frac{1}{8}\)th of its initial amount 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 initial amount of radioactive substance be 1 and haff-life is T.
According to question,
3T = 18000 or, T = \(\frac{18000}{3}\) = 6000 y
Example 4.
An accident in a 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 = 100d
Example 5.
Half-life of thorium is 1.5 × 1010 y. How much time needed for 20% of thorium to disintegrate?
Solution:
Let initial mass of thorium = N0
If in time t 20% of the thorium is disintegrated then,
the amount of thorium that disintegrates
= N0 × \(\frac{20}{100}\) = 0.2 N0
Amount of thorium left,
N = N0 – 0.2N0 = 0.8N0
Now, N = N0e-λt or, eλt = \(\frac{N_0}{N}\) = \(\frac{N_0}{0.8 N_0}\) = 1.25
∴ λt = loge(1.25) = 0.223
or, \(\frac{0.693}{T}\) ᐧ t = 0.223 [∴ T(half-life) = \(\frac{0.693}{\lambda}\)]
or, t = \(\frac{T}{0.693}\) × 0.223
= \(\frac{1.5 \times 10^{10} \times 223}{693}\) = 0.48 × 1010y (approx.)
Alternative method:
N = 0.8N0
Also N = \(\frac{N_0}{2^{t / T}}\)
or, 0.8 = \(\frac{1}{{ }_2^{t / T}}\) or, 2t/T = 5/4
or, t/T = log25/4 = \(\frac{\log _{10} 5 / 4}{\log _{10} 2}\) = \(\frac{0.0969}{0.3010}\) = 0.322
or, t = 0.322T = 0.322 × 1.5 × 1010y
= 0.48 × 1010y (approx.)
Example 6.
Half-life of radium is 1500 y. In how many years will of 1 f of pure radium reduce by 1 mg?
Solution:
Let time in which 1g radium will reduce by 1 mg = t
So, remaining mass of radium = 1 – 0.001 = 0.999 g
Now, assuming initial mass is N0, and in time t mass becomes N then,
\(\frac{N}{N_0}\) = \(\frac{0.999}{1}\) = 0.999
Again, N = N0e-λt
or, eλt = \(\frac{N_0}{N}\) = \(\frac{1}{0.999}\) = 1.001 (approx.)
∴ λt = loge(1.001) = 0.001 (approx.)
or, \(\frac{0.693}{T}\) ᐧ t = 0.001 [∵ half-life, T = \(\frac{0.693}{\lambda}\)]
or, t = \(\frac{T}{0.693}\) × 0.001 = \(\frac{1500}{693}\)y = 2.16 y (approx.)
Example 7.
State the law of radioactive decay. Three fourth of a radioactive sample decays in \(\frac{3}{4}\)s. What is the half-life of the sample?
Solution:
The rate of decay of a radioactive sample with respect to 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 N0 be the number of atoms of a certain radioactive element initially, and N be its number after a time t,
then
N = N0e-λt (where λ = radioactive decay constant)
Given, \(\frac{3}{4}\)th of the sample decays in \(\frac{3}{4}\)s.
So, 2T = \(\frac{3}{4}\)s or, T = \(\frac{3}{8}\)s
Example 8.
A radioactive isotope X with half-life 1.5 × 109 y decays into a stable nucleus Y. A rock sample contains both elements X and Y in ratio 1 : 15. Find the age of the rock. [WBJEE ‘11]
Solution:
X → Y (stable)
Let the quantity of X and Y in the sample be Nx and Ny respectively.