$$\eqalign{ & \Delta E = - 2.0 \times {10^{ - 18}} \times \left( {\frac{1}{{{2^2}}} - \frac{1}{{{1^2}}}} \right) \cr & = - 2.0 \times {10^{ - 18}} \times \frac{{ - 3}}{4} \cr & = 1.5 \times {10^{ - 18}} \cr & \Delta E = \frac{{hc}}{\lambda } \cr & \lambda = \frac{{hc}}{{\Delta E}} \cr & = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{1.5 \times {{10}^{ - 18}}}} \cr & = 1.325 \times {10^{ - 7}}m \cr} $$

$$\eqalign{ & {\text{Given, Planck's constant,}} \cr & h = 6.63 \times {10^{ - 34}}Js \cr & {\text{Speed of light,}}\,\,c = 3 \times {10^{17}}nm\,{s^{ - 1}} \cr & {\text{Frequency of quantam light}} \cr & \nu = 6 \times {10^{15}}{s^{ - 1}} \cr & {\text{Wavelength,}}\,\,\,\lambda = ? \cr & {\text{We know that,}}\,\,\,\nu = \frac{c}{\lambda } \cr & {\text{or,}}\,\,\lambda = \frac{c}{\nu } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3 \times {{10}^{17}}}}{{6 \times {{10}^{15}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 0.5 \times {10^2}\,nm \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 50\,nm \cr} $$

The difference between the energy of adjacent energy levels decreases on moving away from nucleus.

2^nd excited state will be the 3^rd energy level.
$${E_n} = \frac{{13.6}}{{{n^2}}}eV\,\,{\text{or}}\,\,E = \frac{{13.6}}{9}eV$$      $$ = 1.51\,eV.$$

$$\eqalign{ & _{37}Rb = {\,_{36}}\left[ {Kr} \right]5{s^1} \cr & {\text{Its valence electron is}}\,\,5{s^1}. \cr & n = 5 \cr & l = 0\,\,\,\left( {{\text{For }}s{\text{ - orbital}}} \right) \cr & m = 0\,\,\left( {{\text{As }}m = - l\,{\text{to}}\, + l} \right) \cr & s = + \frac{1}{2} \cr} $$

$$\eqalign{ & {\text{Total}}\,{\text{energy}} = \frac{{ - 13.6{Z^2}}}{{{n^2}}}eV \cr & {\text{where}}\,n = 2,3,4\,.... \cr & {\text{Putting}}\,n = 2 \cr & {E_T} = \frac{{ - 13.6}}{4} \cr & = - 3.4\,eV \cr} $$

Energy required to break one mole of $$Cl - Cl$$  bonds in $$C{l_2}$$
$$\eqalign{ & = \frac{{242 \times {{10}^3}}}{{6.023 \times {{10}^{23}}}} = \frac{{hc}}{\lambda } = \frac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{\lambda } \cr & \therefore \,\,\,\lambda = \frac{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8} \times 6.023 \times {{10}^{23}}}}{{242 \times {{10}^8}}} \cr & = 0.4947 \times {10^{ - 6}}m \cr & = 494.7\,nm \cr} $$

No. of radial nodes $$ = n - l - 1$$
For $$3p$$  orbital, $$n = 3,l = 1$$
∴ No. of radial nodes for $$3p$$  orbital
$$\eqalign{ & = 3 - 1 - 1 \cr & = 3 - 2 \cr & = 1 \cr} $$

The lines falling in the visible region comprise Balmer series. Hence the third line from red would be $${n_1} = 2,\,\,{n_2} = 5$$    i.e. $$5 \to 2.$$

$$\eqalign{ & {E_1} = 25\,eV,\,{E_2} = 50\,eV \cr & {E_1} = \frac{{hc}}{{{\lambda _1}}}\,{\text{and}}\,{E_2} = \frac{{hc}}{{{\lambda _2}}} \cr & {\text{or}}\,\,\frac{{{E_1}}}{{{E_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}}\,\,{\text{or}}\,\,\frac{{25}}{{50}} = \frac{{{\lambda _2}}}{{{\lambda _1}}}\,\,{\text{or}}\,\,{\lambda _1} = 2{\lambda _2} \cr} $$