For the Balmer series in the spectrum of $$ $\mathrm{H}$ atom, $\bar{v}={\mathrm{R}}_{\mathrm{H}}\left\{\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right\}$ $\stackrel{}{}{}_{}\left\{\frac{}{{}_{}^{}}\frac{}{{}_{}^{}}\right\}$ the correct statements among (I) to (IV) are :

(I) As wavelength decreases, the lines in the series converge

(II) The integer $\mathrm{n}1$ $$ is equal to 2

(III) The lines of longest wavelength corresponds to $n_2=3$ ${}_{}$

(IV) The ionization energy of hydrogen can be calculated from wave number of these lines

$\mathrm{}\bar{v}=\frac{1}{\lambda }={\mathrm{R}\mathrm{z}}^2?\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$

For $\mathrm{H}$ atom $\mathrm{z}=1$

$\bar{v}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For Balmer series:  ${\mathrm{n}}_1=2$

If ${\mathrm{n}}_2=3\therefore \frac{1}{\lambda }={\mathrm{R}\mathrm{z}}^2\left(\frac{1}{4}-\frac{1}{9}\right)$

$\frac{1}{\lambda }=\mathrm{R}\frac{5}{36}$

${\lambda }_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{36}{5\mathrm{R}}$

$\mathrm{}\stackrel{?}{v}=\frac{1}{?}={\mathrm{R}\mathrm{z}}^2?\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$

For $\mathrm{H}$ atom $\mathrm{z}=1$

$\stackrel{?}{v}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

For Balmer series: ${\mathrm{n}}_1=2$

If ${\mathrm{n}}_2=3?\frac{1}{?}={\mathrm{R}\mathrm{z}}^2\left(\frac{1}{4}-\frac{1}{9}\right)$

$\frac{1}{?}=\mathrm{R}\frac{5}{36}$

${?}_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{36}{5\mathrm{R}}$