• # question_answer Show that the shortest wavelength lines in Lyman, Balmer and Paschen series have their wavelengths in the ratio 1 : 4 : 9. Or Prove that the radius of the nth Bohr orbit of an atom is directly proportional to ${{n}^{2}},$ where n is principal quantum number.

For shortest wavelength of Lyman series ${{n}_{1}}=1,$ ${{n}_{2}}=\infty$ $\frac{1}{{{\lambda }_{LS}}}=R\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{\infty }^{2}}} \right]=R;{{\lambda }_{LS}}=\frac{1}{R}$ For shortest wavelength of Balmer series.             ${{n}_{1}}=2,$ ${{n}_{2}}=\infty$ $\frac{1}{{{\lambda }_{BS}}}=R\left[ \frac{1}{{{2}^{2}}}-\frac{1}{{{\infty }^{2}}} \right]=\frac{R}{4};\,\,{{\lambda }_{BS}}=\frac{4}{R}$ For shortest wavelength of Paschen series             ${{n}_{1}}=3,$ ${{n}_{2}}=\infty$ $\frac{1}{{{\lambda }_{PS}}}=R\left[ \frac{1}{{{3}^{2}}}-\frac{1}{{{\infty }^{2}}} \right]=\frac{R}{9};{{\lambda }_{PS}}=\frac{9}{R}$ Hence, ${{\lambda }_{LS}}:{{\lambda }_{BS}}:{{\lambda }_{PS}}=1:4:9$ Or Radius of nth Bohr's orbit To keep the electron in its orbit, the centripetal force on the electron must be equal to the electrostatic attraction. Therefore, $\frac{m{{v}^{2}}}{r}=\frac{k'Z{{e}^{2}}}{{{r}^{2}}}$ where ,  $k'=\frac{1}{4\pi {{\varepsilon }_{0}}}$ $\Rightarrow$   $r=k'Z{{e}^{2}}/m{{v}^{2}}$                                   ?(i) where, m is the mass of the electron and v its speed in an orbit of radius r. Bohr's quantisation condition for angular momentum is, $mvr=\frac{nh}{2\pi }$ $\Rightarrow r=\frac{nh}{2\pi mv}$                    ?(ii) From Eqs. (i) and (ii), we get             $\frac{k'Z{{e}^{2}}}{m{{v}^{2}}}=\frac{nh}{2\pi mv}\Rightarrow v=\frac{2\pi k'Z{{e}^{2}}}{nh}$ Putting this value of v in Eq. (ii), we get $r=\frac{nh}{2\pi m}\frac{nh}{2\pi k'Z{{e}^{2}}}=\frac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}mk'Z{{e}^{2}}}$ or $r\propto {{n}^{2}}$ Hence proved.