A spectroscopic instrument is capable of resolving two closely...

The wavelength of the lines in the Balmer series is given by the Rydberg formula:

\dfrac{1}{\lambda} = R_H \left(\dfrac{1}{2^2} - \dfrac{1}{n^2} \right)

where $n = 3, 4, 5, \dots$

Differentiating both sides with respect to $n$, we get the wavelength difference between adjacent lines for $\Delta n = 1$:

\dfrac{\Delta \lambda}{\lambda^2} =$R_H$\dfrac{2}{n^3}

$\Delta \lambda =$\dfrac{2 R_H \lambda^2}{n^3}

The resolving power of the instrument is given by:

\dfrac{\lambda}{\Delta \lambda} = \dfrac{n^3}{2 R_H \lambda}

From the Rydberg formula, for large $n$, we can approximate:

\dfrac{1}{\lambda} \approx \dfrac{R_H}{4} \Rightarrow$R_H \lambda \approx 4$

Substituting this into the resolving power expression:

\dfrac{\lambda}{\Delta \lambda} \approx \dfrac{n^3}{2 \times 4} = \dfrac{n^3}{8}

Given that the instrument can resolve lines if \dfrac{\lambda}{\Delta \lambda} \le$8000$, we have:

\dfrac{n^3}{8} \le$8000$

$n^3 \le 64000$

$n \le 40$

The lines in the Balmer series correspond to transitions from $n = 3, 4, 5, \dots, 40$.

The total number of lines that can be resolved is:

$40 - 3 + 1 = 38$

Answer: 38