Physics - Modern Physics Question with Solution | TestHub

$\frac{1}{{\lambda }_{min}}=R\left(1\right)=R;n=\infty$ to $1$

$\frac{1}{{\lambda }_{max}}=R\left\{1-\frac{1}{4}\right\}=\frac{3R}{4}$ ; $n=2$ to $1$

$\Rightarrow$ ${\lambda }_{max}-{\lambda }_{min}=\frac{4}{3R}-\frac{1}{R}$

$304=\frac{1}{3R}$ ......$\left(a\right)$

Paschen : $Y_{{\lambda }_{min}^{\prime }}=R\left(\frac{1}{9}\right)$

and $Y_{{\lambda }_{max}^{\prime }}=R\left(\frac{1}{9}-\frac{1}{16}\right)=\frac{7R}{16\times 9}$

${\left({\lambda }^{{}^{\prime }}\right)}_{max}-{\left({\lambda }^{{}^{\prime }}\right)}_{min}=\frac{16\times 9}{7R}-\frac{9}{R}=\frac{81}{7R}$ ...$\left(b\right)$

$\frac{\left(b\right)}{\left(a\right)}=\frac{x}{3\times 304}=\frac{81}{7}$

$\Rightarrow$ $x=10553.14$