(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. $N = N_0 e^{-\lambda t}$ , where symbols have their usual meaningsLet us consider a radioactive substance having $N_0$ atoms initially at time $(t=0)$ .After time $(t)$ , no. of atoms left undecayed be N.If $dN$ is the no. of atoms decayed in time $dt$ , then according to radioactive decay law: $-\frac{dN}{dt} \alpha N$ $\text{Or, } -\frac{dN}{dt} = \lambda N$ .......(i) Here, $\lambda$ is decay constant and negative sign indicates that a radioactive sample goes on decreasing with time.Equation (i) can also be written as $\frac{dN}{N} = -\lambda dt$Integrating $\ln N = -\lambda t + K$ ........(ii)Here, $K$ is constant of integration $\text{At } t=0, N = N_0$ $\therefore K = \ln N_0$Substituting $K$ in equation (ii), $\ln N = -\lambda t + \ln N_0$ $\ln \frac{N}{N_0} = -\lambda t$Or, $\frac{N}{N_0} = e^{-\lambda t}$ $\therefore N = N_0 e^{-\lambda t}$ (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, $\frac{N}{N_0} = \left( \frac{1}{2} \right)^5 = \frac{1}{32} =$ 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 $\frac{h}{2\pi}$, where $h$ 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 $h v$, which is equal to the difference of the energies of the electron in the two orbits. Expression for Bohr radius : Let us consider $m = \text{mass of an electron}$ $r = \text{radius of the circular orbit in which the}$ $\text{electron is revolving}$ $v = \text{speed of electron}$ $-e = \text{charge of electron}$ $\text{From } 1^{\text{st}} \text{ postulate}$ $\text{Centripetal force} = \text{Electrostatic force}$ $\frac{m v^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2}$ $\therefore v^2 = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{m r}$ .......(1) From $2^{\text{nd}}$ postulate $m v r = \frac{n h}{2\pi}$ Or, $v = \frac{n h}{2 \pi m r}$ Or, $v^2 = \frac{n^2 h^2}{4 \pi^2 m^2 r^2}$ .......(2) Comparing eqns (1) and (2), $\frac{1}{4\pi\varepsilon_0} \frac{e^2}{m r} = \frac{n^2 h^2}{4\pi^2 m^2 r^2}$$\text{Or, } \frac{1}{4\pi\varepsilon_0} \frac{e^2}{m r} = \frac{n^2 h^2}{4\pi^2 m^2 r^2}$$\therefore \text{Bohr radius} = r = \frac{\varepsilon_0 n^2 h^2}{\pi m e^2}$ (b) Shortest wavelength in Balmer series:$\frac{1}{\lambda_S} = R \left( \frac{1}{2^2} - \frac{1}{\infty} \right)$$\therefore \lambda_S = \frac{4}{R}$Longest wavelength in Balmer series: $\therefore \frac{1}{\lambda_L} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$$\therefore \lambda_L = \frac{36}{5 R}$So, $\frac{\lambda_L}{\lambda_S} = \frac{ \frac{36}{5R} }{ \frac{4}{R} } = \frac{9}{5}$