Question

(a) Deduce the expression, $N=N0e^{-\lambda }t$, 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

Let $N_o=$ Total number of atoms present originally in a sample at time $t=0$

$N=$ Total number of atoms left undecayed in the sample at time $t$

$dN=$ A small number of atoms that disintegrate in a small interval of time $dt$

Rate of disintegration of the element, $R=-\frac{dN}{dt}=\lambda N\lambda =disintegrationconstant$

$\frac{dN}{N}=-\lambda dt$

Integrate both side:

$\int \frac{dN}{N}=\int -\lambda dt{\log }_{e}N=-\lambda t+C$

$C=$ constant of integration $t=0$

$N=N_o$

${\log }_{e}N=-\lambda \times 0+C\Rightarrow C={\log }_e{N}_o{\log }_{e}N=\lambda t+{\log }_e{N}_o{\log }_e\frac{N}{N_o}=-\lambda tN=N_0{e}^{-\lambda t}$

(B).

(i). The basic nuclear process underlying this $+\beta$ decay.

$p\to n+e^{+}+v$

The reaction is: ${}_{11}^{22}Na\to {\beta }^{+}{+}_{10}^{22}Ne+v$

(ii) It is an isobar same mass number but different atomic number.