Question

(a) Deduce the expression,\[N = {N_0}{e^{ - \lambda t}}\] for the law of radioactive decay.
(b)
(i) Write symbolically the process expressing the \[\beta + \]decay of \[{}_{11}^{22}Na\]. Also write the basic nuclear process underlying this decay.
(ii) Is the nucleus formed in the decay of the nucleus\[{}_{11}^{22}Na\], an isotope or isobar?

Answer 1

Hint: In a radioactive decay when a material undergoes either \[\alpha \]or \[\beta \] or\[\gamma \]decay the number of nuclei undergoing the decay per unit time is directly proportional to the total number of nuclei in the given sample materials.

Complete step by step answer:
(a)Let
The total number of atoms present originally in a sample at the time t=0 be \[{N_o}\]
The total number of atoms left un-decayed in the sample at time \[t\] be\[N\]
The small number of atom that disintegrate in a small interval of time \[dt\] be \[dN\]
Now according to radioactive decay we know the rate of decay is proportional to amount of nuclei written as
\[
   - \dfrac{{dN}}{{dt}} \propto N \\
   \Rightarrow - \dfrac{{dN}}{{dt}} = \lambda N - - (i) \\
 \]
Where \[\lambda \]is the disintegration constant
Now integrate equation (i)
\[
  \int {\dfrac{{dN}}{N}} = - \lambda \int {dt} \\
   \Rightarrow \log N = - \lambda t + C - - (ii) \\
 \]
Where C is the integration constant total number of atoms present originally in a sample is \[{N_o}\]
Now since at time\[t = 0\] total number of atoms present originally in a sample is \[{N_o}\]
So can write equation (ii) as
\[
  {\log _e}{N_0} = - \lambda \times 0 + C \\
   \Rightarrow C = {\log _e}{N_0} \\
 \]
Hence by substituting the value of C in equation (ii) we get
\[
  {\log _e}N - {\log _e}{N_0} = - \lambda t \\
   \Rightarrow {\log _e}\dfrac{N}{{{N_0}}} = - \lambda t \\
 \]
Hence by taking inverse integration we get
\[
  \dfrac{N}{{{N_0}}} = {e^{ - \lambda t}} \\
   \Rightarrow N = {N_0}{e^{ - \lambda t}} \\
 \]
Hence, the radioactive decay \[N = {N_0}{e^{ - \lambda t}}\]
(b)
(i) Now we know the basic nuclear process underlying the \[\beta + \]decay is \[p \to n + {e^ + } + v\], so the reaction can be written as \[{}_{11}^{22}Na \to {\beta ^ + } + {}_{10}^{22}Ne + v\]
(ii) The nucleus formed in the decay of the nucleus \[{}_{11}^{22}Na\] is \[{}_{10}^{22}Na\] and since nucleus formed has same mass number but different atomic number then it is isobar of ${}_{11}^{22}Na$ .

Note:
It is very important to note that isotopes and isobars are two completely different chemical terms.
Isotopes are atomic structures of the same elements with different mass numbers while on the other hand isobars are different chemical elements having the same atomic mass.