In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is $304 \stackrel{\mathrm{o}}{\mathrm{A}}$. The corresponding difference for the Paschen series in A is 10x. Then the value of x is _____

Form Bohr’s formula for hydrogen atom, $\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right),R=1.097\times {10}^7{m}^{-1}$

For Lyman series: 

$\frac{1}{{\lambda }_{\min }}=R\left(1\right)=R\because n_2=\infty and{n}_1=1$

$\frac{1}{{\lambda }_{\max }}=R\left\{1-\frac{1}{4}\right\}=\frac{3R}{4}\because n_1=2,n_1=1$

$\therefore {\lambda }_{\max }-{\lambda }_{\min .}=\frac{4}{3R}-\frac{1}{R}=\frac{1}{3R}=304\left(given\right)$

For paschen series: 

$\lambda {\text{'}}_{\min }=R\left(\frac{1}{9}\right)and\lambda {\text{'}}_{\max }=R\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7R}{16\times 9}$

$\lambda {\text{'}}_{\min }-\lambda {\text{'}}_{\max }=\frac{16\times 9}{7R}-\frac{9}{R}=\frac{81}{7R}$

$or,\lambda {\text{'}}_{\max }=\lambda {\text{'}}_{\min }=\frac{81}{7R}=\frac{81}{7\times 3R}=\frac{81\times 3}{7}\times 304$

$\left(\because \frac{1}{3R}=304A\right)$

$\therefore$ For Pachen series, $\lambda {\text{'}}_{\max }=\lambda {\text{'}}_{\min }=10553.14$