• # question_answer The wavelength of the first line in Balmer series in the hydrogen spectrum is $\lambda .$What is the wavelength of the second line. A) $\frac{20\lambda }{27}$                               B) $\frac{3\lambda }{16}$                     C)   $\frac{5\lambda }{36}$                     D)   $\frac{3\lambda }{4}$

The first line corresponds to $n=3.$Therefore, $\frac{1}{{{\lambda }_{1}}}=R\left( \frac{1}{4}-\frac{1}{{{3}^{2}}} \right)=R\left( \frac{1}{4}-\frac{1}{9} \right)=\frac{5R}{36}$or ${{\lambda }_{1}}=\frac{36}{5R}.$ For the second line $n=4$ $\therefore$    $\frac{1}{{{\lambda }_{2}}}=R\left( \frac{1}{4}-\frac{1}{{{4}^{2}}} \right)\,\,\Rightarrow {{\lambda }_{2}}=\frac{16}{3R}$ $\therefore$$\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{16}{3R}.\frac{5R}{36}=\frac{20}{27}$or ${{\lambda }_{2}}=\frac{20}{27}\lambda$ Hence, the correction option is (a).