(a) According to Bohr's postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge\\\\(+e\\\\). For an electron moving with a uniform speed in a circular orbit on a given radius, the centripetal force is provided by the coulomb force of attraction between the electron and the nucleus.Therefore,\\\\(\\frac{mv^2}{r}=\\frac{1(Ze)(e)}{4\\pi \\epsilon_0 r^2}\\\\)\\\\(\\implies\\\\) \\\\(mv^2=\\frac{1}{4\\pi \\epsilon_0 } \\frac{Ze^2}{r}\\\\) ....(1)So,Kinetic Energy, \\\\( K.E=\\frac12 mv^2\\\\)\\\\(K.E=\\frac{1}{4\\pi \\epsilon_0} \\frac{Ze^2}{r}\\\\)Potential Energy is given by, \\\\(P. E=\\frac{1}{4\\pi \\epsilon_0} \\frac{(Ze)(-e)}{r}\\\\)Therefore total energy is given by, \\\\(E=K.E+P.E\\\\)\\\\(E=\\frac{1}{4 \\pi \\epsilon_0}\\frac{Ze^2}{2r}+(-\\frac{1}{4\\pi \\epsilon_0} \\frac{Z e^2}{r})\\\\)\\\\(E=-\\frac{1}{4 \\pi \\epsilon_0} \\frac{Ze^2}{2r}\\\\)For, \\\\(n^{th}\\\\) orbit, \\\\(E\\\\) can be written as \\\\(E_n\\\\),\\\\(E_n=-\\frac{1}{4\\pi \\epsilon_0} \\frac{Ze^2}{2r_n}\\\\) ......(2)Now, using Bohr's postulate for quantization of angular momentum, we have\\\\(mvr=\\frac{nh}{2 \\pi}\\\\)\\\\(\\implies\\\\) \\\\(v= \\frac{nh}{2 \\pi mr}\\\\)Putting the values of \\\\(v\\\\) in equation (1), we get\\\\(\\frac{m}{r}\\\\) \\\\((\\frac{nh}{2 \\pi mr})^2=\\frac{1}{4 \\pi \\epsilon_0} \\frac{Ze^2}{r^2}\\\\)\\\\(=-\\frac{mZ^2 e^4}{8 \\epsilon_0 h^2 n^2}\\\\)\\\\(\\implies\\\\) \\\\(\\frac{\\epsilon_0 h^2 n^2}{\\pi m Z e^2}\\\\)Now putting a]value of \\\\(r_n\\\\) in equation (2), we get\\\\(E_n=-\\frac{1}{4 \\pi \\epsilon_0} \\frac{Ze^2}{2(\\frac{\\epsilon_0 h^2 n^2}{ \\pi m})}\\\\)\\\\(=-\\frac{mZ^2e^4}{8 \\epsilon_0 h^2 n^2}\\\\)\\\\(\\implies\\\\) \\\\(E_n=-\\frac{Z^2 Rhc}{n^2}\\\\), where \\\\(R=\\frac{m e^4}{8 \\epsilon_0^2 c h^3}\\\\)where R is the Rydberg constant.(b) Negative sign shows that electron remain bound with the nucleus.(c) If electron jumps from \\\\(n_i=4,5,6 \\\\) to \\\\(n_f=3\\\\), the energy of the line spectra\\\\(\ riangle E = \\frac{me^4}{8 \\epsilon_0^2 h^2} (\\frac{1}{n_f^2}-\\frac{1}{n_i^2})\\\\)