(a) Radioactive decay law : The rate of disintegration of a radioactive substance at an instant is directly proportional to the number of nuclei in the Radioactive substance at that time.
$\mathrm{N}={\mathrm{N}}_{\mathrm{o}}{\mathit{e}}^{-\mathit{\lambda }\mathit{t}}$ , where symbols have their usual meanings
Let us consider a radioactive substance having ${\mathrm{N}}_{\mathrm{o}}$ atoms initially at time $\left(\mathit{t}=0\right)$ .
After time $\left(\mathit{t}\right)$ , no. of atoms left undecayed be N.
If $\mathit{d}\mathrm{N}$ is the no. of atoms decayed in time $\mathit{d}\mathrm{t}$ , then according to radioactive decay law:
$-\frac{\mathit{d}\mathrm{N}}{\mathit{d}\mathit{t}}\mathit{\alpha }\mathrm{N}$
.......(i)
Here, $\mathit{\lambda }$ is decay constant and negative sign indicates that a radioactive sample goes on decreasing with time.
Equation (i) can also be written as
$\frac{\mathit{d}\mathrm{N}}{\mathrm{N}}=-\mathit{\lambda }\mathrm{dt}$
Integrating
$\ln \mathrm{N}=-\mathit{\lambda }\mathit{t}+\mathrm{K}$ ........(ii)
Here, $\mathrm{K}$ is constant of integration

$\therefore \mathrm{K}=\ln {\mathrm{N}}_0$
Substituting $\mathrm{K}$ in equation (ii),
$\ln \mathrm{N}=-\mathit{\lambda }\mathit{t}+\ln {\mathrm{N}}_0$
$\ln \frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{o}}}=-\mathit{\lambda }\mathit{t}$
Or, $\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{o}}}={\mathit{e}}^{-\mathit{\lambda }\mathit{t}}$
$\therefore \mathrm{N}={\mathrm{N}}_0{\mathrm{e}}^{-\mathit{\lambda }\mathit{t}}$
(b) Half-life $=($ Mean life $)\times \ln 2$
Or, $4.5\times {10}^9$ years $=($ mean life $)\times \ln 2$
Or, Mean life $=\left(\frac{4.5\times {10}^9}{\ln 2}\right)$ years
Or, Mean life $=\left(\frac{4.5\times {10}^9}{0.693}\right)$ years $=6.5\times {10}^9$ years
(c) No. of half lives $=5$
So, $\frac{\mathrm{N}}{{\mathrm{N}}_{\mathrm{o}}}={\left(\frac{1}{2}\right)}^5=\frac{1}{32}=$ fraction of mass of the radioactive substance left undecayed
So, fraction of mass decayed $=1-\frac{1}{32}=\frac{31}{32}$
OR
(a) Postulates of Bohr Model of Hydrogen atom:
Postulate - I: The electrons revolve in a circular orbit around the nucleus. The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons provide necessary centripetal force for circular motion.
Postulate - II: The electrons can revolve only in certain selected orbits in which angular momentum of electrons is equal to the integral multiple $\frac{\mathit{h}}{2\mathit{\pi }}$, where $\mathit{h}$ is Planck's constant. These orbits are known as stationary or permissible orbits. The electrons do not radiate energy while revolving in theses orbits.
Postulate - III: When an electrons jumps from higher energy orbit to lower energy orbit, energy is radiated in the form of a quantum or photon of energy $\mathit{h}\mathit{v}$, which is equal to the difference of the energies of the electron in the two orbits.
Expression for Bohr radius :
Let us consider







$\frac{\mathit{m}{\mathit{v}}^2}{\mathit{r}}=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{{\mathit{e}}^2}{{\mathit{r}}^2}$
$\therefore {\mathit{v}}^2=\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{{\mathit{e}}^2}{\mathit{m}\mathit{r}}$ .......(1)
From postulate
$\mathit{m}\mathit{v}\mathit{r}=\frac{\mathit{n}\mathit{h}}{2\mathit{\pi }}$
Or, $\mathit{v}=\frac{\mathit{n}\mathit{h}}{2\mathit{\pi }\mathit{m}\mathit{r}}$
Or, ${\mathit{v}}^2=\frac{{\mathit{n}}^2{\mathit{h}}^2}{4{\mathit{\pi }}^2{\mathit{m}}^2{\mathit{r}}^2}$ .......(2)
Comparing eqns (1) and (2),
$\frac{1}{4\mathit{\pi }{\mathit{\epsilon }}_0}\frac{{\mathit{e}}^2}{\mathit{m}\mathit{r}}=\frac{{\mathit{n}}^2{\mathit{h}}^2}{4{\mathit{\pi }}^2{\mathit{m}}^2{\mathit{r}}^2}$


(b) Shortest wavelength in Balmer series:
$\frac{1}{{\mathit{\lambda }}_{\mathit{S}}}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{\mathit{\infty }}\right)$
$\therefore {\mathit{\lambda }}_{\mathit{S}}=\frac{4}{\mathrm{R}}$
Longest wavelength in Balmer series:
$\therefore \frac{1}{{\mathit{\lambda }}_{\mathit{L}}}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$
$\therefore {\mathit{\lambda }}_{\mathit{L}}=\frac{36}{5\mathrm{R}}$
So, $\frac{{\mathit{\lambda }}_{\mathit{L}}}{{\mathit{\lambda }}_{\mathit{S}}}=\frac{\frac{36}{5\mathrm{R}}}{\frac{4}{\mathrm{R}}}=\frac{9}{5}$