Solution: Spectral Line Resolution
Key Concept: The resolving power $R$ of a spectrometer is defined as $\frac{\lambda}{|\Delta \lambda|}$. For the Balmer series, the wavelength $\lambda$ for a transition from $n \to 2$ is given by the Rydberg formula. As $n$ increases, the lines get closer together ($\Delta \lambda$ decreases). We need to find the highest $n$ for which the spectral separation is still resolvable.
Step 1: Wavelength Formula
For the Balmer series ($n_1=2, n_2=n$), the wavelength $\lambda$ is: $$ \frac{1}{\lambda} = R \left( \frac{1}{2^2} – \frac{1}{n^2} \right) = R \left( \frac{1}{4} – \frac{1}{n^2} \right) $$ For large $n$, $\frac{1}{\lambda} \approx \frac{R}{4}$, so $\lambda \approx \frac{4}{R}$.
For the Balmer series ($n_1=2, n_2=n$), the wavelength $\lambda$ is: $$ \frac{1}{\lambda} = R \left( \frac{1}{2^2} – \frac{1}{n^2} \right) = R \left( \frac{1}{4} – \frac{1}{n^2} \right) $$ For large $n$, $\frac{1}{\lambda} \approx \frac{R}{4}$, so $\lambda \approx \frac{4}{R}$.
Step 2: Differentiate to find separation $\Delta \lambda$
Differentiating the Rydberg equation with respect to $n$: $$ -\frac{1}{\lambda^2} d\lambda = R \left( 0 – (-2n^{-3}) \right) dn $$ $$ -\frac{1}{\lambda^2} d\lambda = \frac{2R}{n^3} dn $$ Considering the magnitude of separation between adjacent lines (where $\Delta n = 1$): $$ |\Delta \lambda| \approx \lambda^2 \frac{2R}{n^3} $$
Differentiating the Rydberg equation with respect to $n$: $$ -\frac{1}{\lambda^2} d\lambda = R \left( 0 – (-2n^{-3}) \right) dn $$ $$ -\frac{1}{\lambda^2} d\lambda = \frac{2R}{n^3} dn $$ Considering the magnitude of separation between adjacent lines (where $\Delta n = 1$): $$ |\Delta \lambda| \approx \lambda^2 \frac{2R}{n^3} $$
Step 3: Apply Resolving Power Condition
The condition for resolution is: $$ \frac{\lambda}{|\Delta \lambda|} \le \text{Resolving Power} $$ Substituting $|\Delta \lambda|$: $$ \frac{\lambda}{\lambda^2 \frac{2R}{n^3}} = \frac{n^3}{2R\lambda} $$ Since $\lambda \approx \frac{4}{R}$, substitute $\lambda$: $$ \text{Resolving Power} \approx \frac{n^3}{2R(4/R)} = \frac{n^3}{8} $$
The condition for resolution is: $$ \frac{\lambda}{|\Delta \lambda|} \le \text{Resolving Power} $$ Substituting $|\Delta \lambda|$: $$ \frac{\lambda}{\lambda^2 \frac{2R}{n^3}} = \frac{n^3}{2R\lambda} $$ Since $\lambda \approx \frac{4}{R}$, substitute $\lambda$: $$ \text{Resolving Power} \approx \frac{n^3}{2R(4/R)} = \frac{n^3}{8} $$
Step 4: Calculation
Given Resolving Power $= 8 \times 10^3 = 8000$. $$ \frac{n^3}{8} = 8000 $$ $$ n^3 = 64000 $$ $$ n = \sqrt[3]{64000} = 40 $$
Given Resolving Power $= 8 \times 10^3 = 8000$. $$ \frac{n^3}{8} = 8000 $$ $$ n^3 = 64000 $$ $$ n = \sqrt[3]{64000} = 40 $$
Correct Option: (b) 40