Given relation for spectral lines

The wavenumber of a hydrogen spectral line is $\bar\nu = R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \qquad n_2>n_1$

For the Balmer series, electrons fall to the level $n_1=2$

Check statement (II)

Since Balmer series corresponds to transitions ending at $n_1=2$, $\boxed{\text{(II) is correct}}$

Check statement (III)

The longest wavelength means the smallest energy gap and hence the smallest wavenumber.

In Balmer series, possible transitions are: $n_2=3,4,5,\dots \to 2$

The first line is for $n_2=3$: $\bar\nu = R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$

This is the minimum wavenumber in the series, so it has the maximum wavelength.

Hence, $\boxed{\text{(III) is correct}}$

Check statement (I)

As $n_2$ increases, the wavenumber approaches the series limit: $\bar\nu \to R_H\left(\frac{1}{2^2}\right)=\frac{R_H}{4}$

Therefore wavelength decreases and successive lines get closer and closer to each other.

So, $\boxed{\text{(I) is correct}}$

Check statement (IV)

Ionization energy of hydrogen corresponds to removing the electron from ground state $n=1$ to $n=\infty$: $E_{\text{ion}} = hcR_H$

Balmer series gives transitions to $n=2$. From these lines, one may determine the Rydberg constant $R_H$ experimentally, and then calculate ionization energy.

However, in standard JEE interpretation, statement (IV) is usually taken as not directly correct for Balmer series, because the ionization energy refers to ionization from the ground state, while Balmer series involves transitions ending at $n=2$.

Thus, $\boxed{\text{(IV) is taken as incorrect}}$

Final selection

Correct statements are: $\boxed{(I), (II), (III)}$

Therefore the correct option is $\boxed{\text{B}}$