For emission line of atomic hydrogen from ni = 8 to nf = n, the plot of wave number () against (1n2) will be: (The…

🧠 Transforming the Formula Graphing in physical chemistry is about turning a complex equation into a linear one ($y = mx + c$). We need to take the Rydberg emission formula and isolate our variables for the axes provided.

🗺️ The Algebraic Linearization

1. The Core Formula: $\bar{\nu} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
2. Plugging the Constants: Given transition is from $n_i = 8$ to $n_f = n$. $\bar{\nu} = R_H \left( \frac{1}{n^2} - \frac{1}{8^2} \right) = R_H \left( \frac{1}{n^2} - \frac{1}{64} \right)$
3. Linear Form: $\bar{\nu} = \underbrace{R_H}_{m} \cdot \underbrace{\left(\frac{1}{n^2}\right)}_{x} - \underbrace{\frac{R_H}{64}}_{c}$ Here, the slope ($m$) is clearly the Rydberg constant ($R_H$). Since $R_H$ is a positive constant, the slope is positive.

⚡ The "Initial is Fixed" Trick Notice that $n_i$ is fixed at a high value ($8$) and $n$ is the variable. If $n$ increases, the energy gap decreases. But we are plotting against $1/n^2$. As $1/n^2$ increases, the wave number $\bar{\nu}$ also increases with a constant multiplicative factor $R_H$. This confirms a positive linear relationship.

⚠️ Common Traps The sign of the slope! If the graph was against $1/n_i^2$ with fixed $n_f$, the slope would be $-R_H$. Always check which term (initial or final) is the variable on your x-axis.

$\boxed{\text{Answer: (c)}}$