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(C) For an emission line,$n_f < n_i$. The Rydberg formula for the wave number is $\bar{
u} = R_H Z^2 \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} ight]

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Linear with slope $R_H$

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(C) For an emission line,$n_f The Rydberg formula for the wave number is $\bar{
u} = R_H Z^2 \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} ight]$. For atomic hydrogen,$Z = 1$,$n_i = 8$,and $n_f = n$. Substituting these values: $\bar{
u} = R_H \left[ \frac{1}{n^2} - \frac{1}{8^2} ight] = R_H \left( \frac{1}{n^2} ight) - \frac{R_H}{64}$. Comparing this with the equation of a straight line $y = mx + c$,where $y = \bar{
u}$ and $x = \frac{1}{n^2}$: The slope $m = R_H$ and the intercept $c = -\frac{R_H}{64}$. Thus,the plot is linear with a slope of $R_H$.

For an 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).

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