For emission line of atomic hydrogen from $$n_i = 8$$ to $$n_f = n$$, the plot of wave number $$\bar{\nu}$$ against $$\frac{1}{n^2}$$ will be: (The Rydberg constant, $$R_H$$ is in wave number unit)

We begin with the well-known Rydberg formula for the spectrum of atomic hydrogen. In terms of wave number (the reciprocal of wavelength) the formula is stated as

$$\bar{\nu}=R_H\left(\dfrac{1}{n_f^{\,2}}-\dfrac{1}{n_i^{\,2}}\right)$$

Here $$n_i$$ is the principal quantum number of the initial (higher) level and $$n_f$$ is that of the final (lower) level. For the emission line mentioned in the problem we have the fixed initial level $$n_i = 8$$ and the variable final level $$n_f = n$$, where obviously $$n<8$$ because energy is being released.

Substituting $$n_i = 8$$ into the formula we get

$$\bar{\nu}=R_H\left(\dfrac{1}{n^{2}}-\dfrac{1}{8^{2}}\right)$$

Since $$8^{2}=64$$, the expression simplifies step by step as follows:

$$\bar{\nu}=R_H\left(\dfrac{1}{n^{2}}-\dfrac{1}{64}\right)$$

$$\bar{\nu}=R_H\left(\dfrac{1}{n^{2}}\right)-R_H\left(\dfrac{1}{64}\right)$$

$$\bar{\nu}=R_H\left(\dfrac{1}{n^{2}}\right)-\dfrac{R_H}{64}$$

Let us now introduce the variable

$$x=\dfrac{1}{n^{2}}$$

Re-writing the result in terms of this new variable gives

$$\bar{\nu}=R_H\,x-\dfrac{R_H}{64}$$

We can see clearly that the right-hand side is of the form $$y=mx+c$$, where the dependent variable is $$y=\bar{\nu}$$, the independent variable is $$x=\dfrac{1}{n^{2}}$$, the slope (or gradient) is $$m=R_H$$, and the y-intercept is $$c=-\dfrac{R_H}{64}$$.

Thus, the graph of $$\bar{\nu}$$ versus $$\dfrac{1}{n^{2}}$$ is a straight line (i.e. linear), and the slope of this straight line is positive and equal to $$R_H$$.

Hence, the correct answer is Option C.