`color{blue} ✍️`Becquerel discovered radioactivity in 1896 purely by accident. While studying the fluorescence and phosphorescence of compounds irradiated with visible light, Becquerel observed an interesting phenomenon.
`color{blue} ✍️`After illuminating some pieces of uranium-potassium sulphate with visible light, he wrapped them in black paper and separated the package from a photographic plate by a piece of silver.
`color{blue} ✍️`When, after several hours of exposure, the photographic plate was developed, it showed blackening due to something that must have been emitted by the compound and was able to penetrate both black paper and the silver.
`color{blue} ✍️`Experiments performed subsequently showed that radioactivity was a nuclear phenomenon in which an unstable nucleus undergoes a decay. This is referred to as radioactive decay. Three types of radioactive decay occur in nature :
`color{blue} {(i)}` `alpha`-decay in which a helium nucleus `text()_(2)^(4)He` is emitted;
`color{blue} {(ii)}` `beta`-decay in which electrons or positrons (particles with the same mass as electrons, but with a charge exactly opposite to that of electron) are emitted;
`color{blue} {(iii)}` `gamma`-decay in which high energy (hundreds of keV or more) photons are emitted.
Each of these decay will be considered in subsequent sub-sections.
`color{brown}bbul("Law of radioactive decay")`
`color{blue} ✍️`In any radioactive sample, which undergoes `alpha, beta` or g-decay, it is found that the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample. If N is the number of nuclei in the sample and `DeltaN` undergo decay in time `Deltat` then
`color{blue} ✍️` `color{green}{(DeltaN)/(Deltat) proptoN}`
or,
`color{navy}(DeltaN//Deltat = λN,)`
..............(13.10)
`color{blue} ✍️`where `I` is called the radioactive decay constant or disintegration constant. The change in the number of nuclei in the sample* is `dN = – DeltaN` in time `Deltat`.
`color{blue} ✍️`Thus the rate of change of N is (in the limit `Deltat -> 0`)
`color{purple}((Dn)/(dt) = - lamdaN)`
or
`color{purple}((dN)/N =- lamdadt)`
`color{blue} ✍️`Now, integrating both sides of the above equation,we get,
`color{navy}(int_(N0)^(N) (dN)/N = - lamda int_(t_0)^(t) dt)`
..........(13.11)
or,
`color{navy}(ln N − ln N_0 = −lamda(t – t_0))`
............(13.12)
`color{blue} ✍️`Here `N_0` is the number of radioactive nuclei in the sample at some arbitrary time `t_0` and N is the number of radioactive nuclei at any subsequent time t. Setting `t_0 = 0` and rearranging Eq. (13.12) gives us
`color{navy}(In (N)/(N_0) = - lamdat)`
......(13.13)
which gives
`color{navy}(N(t) = N_0 e^(−lamda t))`
....... (13.14)
`color{brown} {"Note,"}` for example, the light bulbs follow no such exponential decay law. If we test 1000 bulbs for their life (time span before they burn out or fuse), we expect that they will ‘decay’ (that is, burn out) at more or less the same time.
`color{blue} ✍️`The decay of radionuclides follows quite a different law, the law of radioactive decay represented by Eq. (13.14).
`color{blue} ✍️`The total decay rate `R` of a sample is the number of nuclei disintegrating per unit time. Suppose in a time interval dt, the decay count measured is `DeltaN.` Then `dN = – DeltaN.`
`color{blue} ✍️`The positive quantity `R` is then defined as
`color{purple}(R = - (dN)/(dt))`
`color{blue} ✍️`Differentiating Eq. (13.14), we get
`color{purple}(R = lamdaN_0e^(-lamdat))`
or
`color{purple}(R= R_0e^(-lamdat))`
`color{blue} ✍️`This is equivalant to the law of radioactivity decay, since you can integrate Eq. (13.15) to get back Eq. (13.14). Clearly, `R_0 = lamdaN_0` is the decay rate at `t = 0`.
`color{blue} ✍️`The decay rate `R` at a certain time t and the number of undecayed nuclei N at the same time are related by
`color{navy}(R = lamdaN)`
.......(13.16)
`color{blue} ✍️`The decay rate of a sample, rather than the number of radioactive nuclei, is a more direct experimentally measurable quantity and is given a specific name: activity. The SI unit for activity is becquerel, named after the discoverer of radioactivity, Henry Becquerel.
`color{blue} ✍️`1 becquerel is simply equal to 1 disintegration or decay per second. There is also another unit named “curie” that is widely used and is related to the SI unit as:
`color{purple}(1 "curie" = 1 Ci = 3.7 × 10^(10) "decays per second")`
`color{purple}(= 3.7 × 10^(10))` Bq
`color{blue} ✍️`Different radionuclides differ greatly in their rate of decay. A common way to characterize this feature is through the notion of half-life.
Half-life of a radionuclide (denoted by `T_(1//2)`) is the time it takes for a sample that has initially, say `N_0` radionuclei to reduce to
`color{purple}(N_0//2).` Putting `color{purple}(N = N_0//2)` and `color{purple}(t = T_(1//2))` in Eq. (13.14), we get
`color{navy}(T_(1//2)= (In2)/(lamda) = (0.693)/(lamda))`
....... (13.17)
`color{blue} ✍️`Clearly if `N_0` reduces to half its value in time `color{green}(T_(1//2), R_0)` will also reduce to half its value in the same time according to Eq. (13.16).
`color{blue} ✍️`Another related measure is the average or mean life t. This again can be obtained from Eq. (13.14).
`color{blue} ✍️`The number of nuclei which decay in the time interval t to `t + Deltat` is `R(t)Deltat = lamdaN_0e^(–lamdat)Deltat`.
Each of them has lived for time t.
`color{blue} ✍️`Thus the total life of all these nuclei would be `tlamdaN_0 e^(-lamdat) Deltat` It is clear that some nuclei may live for a short time while others may live longer. Therefore to obtain the mean life, we have to sum (or integrate) this expression over all times from `0` to `infty` , and divide by the total number `N_0` of nuclei at `t = 0.`
`color{blue} ✍️`Thus, `color{purple}(τ = (lamdaN_0 int_(0)^(infty) te^(-lamdat)dt)/(N_0) = lamdaint_(0)^(infty)te^(-lamdat)dt)`
One can show by performing this integral that
`color{purple}(τ = 1//λ)`
`color{blue} ✍️`We summarise these results with the following:
`color{navy}{T_(1//2)= (In2)/(lamda) = τIn2}`
............ (13.18)
`color{blue} ✍️`Those radioactive elements whose half-life is short compared to the age of the universe (13.7 billion years) are not found in observable quantities in nature today.
`color{blue} ✍️`They have, however, been seen in the laboratory in nuclear reactions. Tritium and plutonium belong to this category.
`color{blue} ✍️`Becquerel discovered radioactivity in 1896 purely by accident. While studying the fluorescence and phosphorescence of compounds irradiated with visible light, Becquerel observed an interesting phenomenon.
`color{blue} ✍️`After illuminating some pieces of uranium-potassium sulphate with visible light, he wrapped them in black paper and separated the package from a photographic plate by a piece of silver.
`color{blue} ✍️`When, after several hours of exposure, the photographic plate was developed, it showed blackening due to something that must have been emitted by the compound and was able to penetrate both black paper and the silver.
`color{blue} ✍️`Experiments performed subsequently showed that radioactivity was a nuclear phenomenon in which an unstable nucleus undergoes a decay. This is referred to as radioactive decay. Three types of radioactive decay occur in nature :
`color{blue} {(i)}` `alpha`-decay in which a helium nucleus `text()_(2)^(4)He` is emitted;
`color{blue} {(ii)}` `beta`-decay in which electrons or positrons (particles with the same mass as electrons, but with a charge exactly opposite to that of electron) are emitted;
`color{blue} {(iii)}` `gamma`-decay in which high energy (hundreds of keV or more) photons are emitted.
Each of these decay will be considered in subsequent sub-sections.
`color{brown}bbul("Law of radioactive decay")`
`color{blue} ✍️`In any radioactive sample, which undergoes `alpha, beta` or g-decay, it is found that the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample. If N is the number of nuclei in the sample and `DeltaN` undergo decay in time `Deltat` then
`color{blue} ✍️` `color{green}{(DeltaN)/(Deltat) proptoN}`
or,
`color{navy}(DeltaN//Deltat = λN,)`
..............(13.10)
`color{blue} ✍️`where `I` is called the radioactive decay constant or disintegration constant. The change in the number of nuclei in the sample* is `dN = – DeltaN` in time `Deltat`.
`color{blue} ✍️`Thus the rate of change of N is (in the limit `Deltat -> 0`)
`color{purple}((Dn)/(dt) = - lamdaN)`
or
`color{purple}((dN)/N =- lamdadt)`
`color{blue} ✍️`Now, integrating both sides of the above equation,we get,
`color{navy}(int_(N0)^(N) (dN)/N = - lamda int_(t_0)^(t) dt)`
..........(13.11)
or,
`color{navy}(ln N − ln N_0 = −lamda(t – t_0))`
............(13.12)
`color{blue} ✍️`Here `N_0` is the number of radioactive nuclei in the sample at some arbitrary time `t_0` and N is the number of radioactive nuclei at any subsequent time t. Setting `t_0 = 0` and rearranging Eq. (13.12) gives us
`color{navy}(In (N)/(N_0) = - lamdat)`
......(13.13)
which gives
`color{navy}(N(t) = N_0 e^(−lamda t))`
....... (13.14)
`color{brown} {"Note,"}` for example, the light bulbs follow no such exponential decay law. If we test 1000 bulbs for their life (time span before they burn out or fuse), we expect that they will ‘decay’ (that is, burn out) at more or less the same time.
`color{blue} ✍️`The decay of radionuclides follows quite a different law, the law of radioactive decay represented by Eq. (13.14).
`color{blue} ✍️`The total decay rate `R` of a sample is the number of nuclei disintegrating per unit time. Suppose in a time interval dt, the decay count measured is `DeltaN.` Then `dN = – DeltaN.`
`color{blue} ✍️`The positive quantity `R` is then defined as
`color{purple}(R = - (dN)/(dt))`
`color{blue} ✍️`Differentiating Eq. (13.14), we get
`color{purple}(R = lamdaN_0e^(-lamdat))`
or
`color{purple}(R= R_0e^(-lamdat))`
`color{blue} ✍️`This is equivalant to the law of radioactivity decay, since you can integrate Eq. (13.15) to get back Eq. (13.14). Clearly, `R_0 = lamdaN_0` is the decay rate at `t = 0`.
`color{blue} ✍️`The decay rate `R` at a certain time t and the number of undecayed nuclei N at the same time are related by
`color{navy}(R = lamdaN)`
.......(13.16)
`color{blue} ✍️`The decay rate of a sample, rather than the number of radioactive nuclei, is a more direct experimentally measurable quantity and is given a specific name: activity. The SI unit for activity is becquerel, named after the discoverer of radioactivity, Henry Becquerel.
`color{blue} ✍️`1 becquerel is simply equal to 1 disintegration or decay per second. There is also another unit named “curie” that is widely used and is related to the SI unit as:
`color{purple}(1 "curie" = 1 Ci = 3.7 × 10^(10) "decays per second")`
`color{purple}(= 3.7 × 10^(10))` Bq
`color{blue} ✍️`Different radionuclides differ greatly in their rate of decay. A common way to characterize this feature is through the notion of half-life.
Half-life of a radionuclide (denoted by `T_(1//2)`) is the time it takes for a sample that has initially, say `N_0` radionuclei to reduce to
`color{purple}(N_0//2).` Putting `color{purple}(N = N_0//2)` and `color{purple}(t = T_(1//2))` in Eq. (13.14), we get
`color{navy}(T_(1//2)= (In2)/(lamda) = (0.693)/(lamda))`
....... (13.17)
`color{blue} ✍️`Clearly if `N_0` reduces to half its value in time `color{green}(T_(1//2), R_0)` will also reduce to half its value in the same time according to Eq. (13.16).
`color{blue} ✍️`Another related measure is the average or mean life t. This again can be obtained from Eq. (13.14).
`color{blue} ✍️`The number of nuclei which decay in the time interval t to `t + Deltat` is `R(t)Deltat = lamdaN_0e^(–lamdat)Deltat`.
Each of them has lived for time t.
`color{blue} ✍️`Thus the total life of all these nuclei would be `tlamdaN_0 e^(-lamdat) Deltat` It is clear that some nuclei may live for a short time while others may live longer. Therefore to obtain the mean life, we have to sum (or integrate) this expression over all times from `0` to `infty` , and divide by the total number `N_0` of nuclei at `t = 0.`
`color{blue} ✍️`Thus, `color{purple}(τ = (lamdaN_0 int_(0)^(infty) te^(-lamdat)dt)/(N_0) = lamdaint_(0)^(infty)te^(-lamdat)dt)`
One can show by performing this integral that
`color{purple}(τ = 1//λ)`
`color{blue} ✍️`We summarise these results with the following:
`color{navy}{T_(1//2)= (In2)/(lamda) = τIn2}`
............ (13.18)
`color{blue} ✍️`Those radioactive elements whose half-life is short compared to the age of the universe (13.7 billion years) are not found in observable quantities in nature today.
`color{blue} ✍️`They have, however, been seen in the laboratory in nuclear reactions. Tritium and plutonium belong to this category.