Radioactivity:
In the beginning of this chapter, it was mentioned that Henry Becquerel discovered radioactivity in 1896. Heavy elements like uranium, thorium, etc. particles or radiations on their own and undergoes a decay. In this process of decay, new atoms (elements) are formed that themselves exhibit radioactivity and this process continues till a stable element is formed.
The discovery of radioactivity by Henry Becquerel was purely accidental. He observed that when uranium salt-crystals or uranium-potassium sulphate were illuminated with visible light, they emitted some invisible radiations that blackened the photograhic plate placed near it which was covered by a light resistant cover. After this, scientists Marie Curie and Pierre Curie discovered the new elements radium and polonium from the pitch blend are which were more radioactive than uranium. Through the experiments, Rutherford showed that radioactive radiations are of three type which he called alpha, beta and gamma rays. The experiments performed later showed that-α (alpha) rays are helium nuclei, β (beta) rays are electrons or positrons and γ (gamma) rays are photons of high energy. Experiments also showed that the radioactivity is a result of decay of unstable nuclei. This way, we can say that radioactivity is a nuclear phenomenon. Some important facts related to radioactivity are as follows :
(i) External conditions like pressure temperature and state of radioactive substance (solid, liquid or gas) has no effect on radioactivity. Radioactivity is also unaffected by the chemical reactions or chemical combinations (like uranium or any of its compounds is radioactive). As there is contribution of only the outermost electrons in a chemical combination. Thus, the electronic configuration of atoms is not related to the radioactivity. Also, α-ravs, β-particles of very high energy or emission of γ-ray photons is not possible from the emission of outer part of atom. Thus, radioactivity is purely a nuclear phenomenon.
(ii) In the radioactive decay of a nuclei, with the laws of conservation of mass, energy, charge linear momentum, angular momentum and the conservation of number of nucleons should be followed.
(iii) A nucleus say X is unstable for the decay of α and β if its mass is more than the sum of its product constituents.
(iv) In the process of radioactive decay, emission energy per atom is the order of few MeV where as it of the order of few eV for chemical reactions.
Half Life:
We know that radioactive elements always disintegrated and as the time passes the number of united nuclei decreases. “Half life of a radioactive sample is the time to reduce the nuclei to half of its initial value.” Half life is denoted by T. Its value for an element is constant and varies for different elements. Half life does not depend on the quantity of substance. It cannot be replaced by physical and chemical effects.
The half life of some elements are given below :
| S. No. |
Symbol |
Half-life |
| 1. |
Uranium (92U238) |
4.5 × 109 year |
| 2. |
Thorium (90Th230) |
8 × 104 year |
| 3. |
Radium (88Ra236) |
1620 year |
| 4. |
Bismuth (83Bi218) |
3 min |
If a radioactive element has a half life of T then after T times it will have 50% of its initial value, after 2T times 25% and after 3T times 12.5%, after 4T time 6.25% remaining.
If the graph plotted between the number of nuclei and time, then the graph is shown in a figure.
Let the number of nuclei in beginning is N0 that is N = 0 at t = 0. So remaining nuclei after a half life (t = T),
Similarly after three half lives (t = nT), remaining nuclear
Nn = N0 \(\left(\frac{1}{2}\right)^{n}\) ……………… (1)
∵ t = nT
Using n = \(\frac{t}{T}\), the value of n can be determined
Relation between Half-life and Decay constant
If the number of neclei in the beginning is N0 then the number of remaining nuclei after time t
N = N0e-λt
Taking log on both sides
loge = logeeλt = λt loge e = λt
or λt = loge 2
or T = \(\frac{\log _{e} 2}{\lambda}\) …………… (2)
With the help of this equation we can find out the value of λ it value of T is known or vice-versa,
∵ loge 2 = 0.6932
∴ T = \(\frac{0.6932}{\lambda}\) …………. (3)
or λ = \(\frac{0.6932}{T}\) …………….. (4)
