# In the line spectra of hydrogen atom,

Question:

In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is $304 A. The corresponding difference for the Paschan series in A is :______ Solution: (10553.14) From Bohr's formula for hydrogen atom,$\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)R=1.097 \times 10^{7} \mathrm{~m}^{-1}$For Lyman series :$\frac{1}{\lambda_{\min .}}=R(1)=R \quad \because n_{2}=\infty$and$n_{1}=1\frac{1}{\lambda_{\max }}=R\left\{1-\frac{1}{4}\right\}=\frac{3 R}{4} \quad \because n_{1}=2, n_{1}=1\therefore \lambda_{\max }-\lambda_{\min .}=\frac{4}{3 R}-\frac{1}{R}=\frac{1}{3 R}=304$(Given) For Paschen series :$\lambda_{\min .}^{\prime}=R\left(\frac{1}{9}\right)$and$\lambda_{\max .}^{\prime}=R\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7 R}{16 \times 9}\lambda_{\max .}^{\prime}-\lambda_{\min .}^{\prime}=\frac{16 \times 9}{7 R}-\frac{9}{R}=\frac{81}{7 R}$or,$\lambda_{\max .}^{\prime}-\lambda_{\min .}^{\prime}=\frac{81}{7 R}=\frac{81 \times 3}{7 \times 3 R}=\frac{81 \times 3}{7} \times 304\left(\because \frac{1}{3 R}=304 Aright)\therefore$For Pachen series,$\lambda_{\max .}^{\prime}-\lambda_{\min .}^{\prime}=10553.14\$