1. Use the Rydberg formula

For hydrogen spectrum, $\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right), \qquad n_2>n_1$ where $R$ is the Rydberg constant.

For a given series, $n_1$ is fixed:

• Lyman: $n_1=1$
• Balmer: $n_1=2$
• Paschen: $n_1=3$

For each series:

• Longest wavelength corresponds to the smallest energy gap, i.e. $n_2=n_1+1$
• Shortest wavelength corresponds to the series limit, i.e. $n_2\to\infty$

2. Check option A: Ratio of longest to shortest wavelength in Balmer series

For Balmer series, $n_1=2$.

• Shortest wavelength: $\frac{1}{\lambda_{\min}}=R\left(\frac{1}{2^2}-0\right)=\frac{R}{4}$ So, $\lambda_{\min}=\frac{4}{R}$

• Longest wavelength: transition $3\to 2$ $\frac{1}{\lambda_{\max}}=R\left(\frac{1}{4}-\frac{1}{9}\right)=R\cdot\frac{5}{36}$ So, $\lambda_{\max}=\frac{36}{5R}$

Therefore, $\frac{\lambda_{\max}}{\lambda_{\min}}=\frac{36}{5R}\cdot\frac{R}{4}=\frac{9}{5}$

So A is correct.

3. Check option B: Overlap between Balmer and Paschen wavelength ranges

Balmer series range

For Balmer ($n_1=2$):

• shortest wavelength: $\lambda_{B,\min}=\frac{4}{R}$
• longest wavelength: $\lambda_{B,\max}=\frac{36}{5R}$

Thus, $\lambda \in \left[\frac{4}{R},\,\frac{36}{5R}\right]$

Paschen series range

For Paschen ($n_1=3$):

• shortest wavelength: $\frac{1}{\lambda_{P,\min}}=R\left(\frac{1}{9}\right) \Rightarrow \lambda_{P,\min}=\frac{9}{R}$
• longest wavelength: transition $4\to 3$ $\frac{1}{\lambda_{P,\max}}=R\left(\frac{1}{9}-\frac{1}{16}\right)=R\cdot\frac{7}{144}$ So, $\lambda_{P,\max}=\frac{144}{7R}$

Thus, $\lambda \in \left[\frac{9}{R},\,\frac{144}{7R}\right]$

Now compare Balmer and Paschen ranges:

• Balmer maximum = $\frac{36}{5R}=\frac{7.2}{R}$
• Paschen minimum = $\frac{9}{R}$

Since $\frac{36}{5R}<\frac{9}{R}$ there is no overlap.

So B is incorrect.

4. Check option C: Formula for wavelengths of Lyman series

For Lyman series, $n_1=1$, $n_2=m$ with $m=2,3,4,\dots$

Then, $\frac{1}{\lambda}=R\left(1-\frac{1}{m^2}\right)$ So, $\lambda=\frac{1}{R\left(1-\frac{1}{m^2}\right)}$ =\frac{1}{R}\cdot \frac{1}{1-\frac{1}{m^2}}$$

The shortest wavelength of Lyman series is for $m\to\infty$: $\lambda_0=\frac{1}{R}$

Hence, $\lambda=\lambda_0\cdot \frac{1}{1-\frac{1}{m^2}}=\lambda_0\cdot \frac{m^2}{m^2-1}$

This is not equal to $\left(1+\frac{1}{m^2}\right)\lambda_0$

So C is incorrect.

5. Check option D: Lyman and Balmer ranges do not overlap

Lyman range

For Lyman ($n_1=1$):

• shortest wavelength: $\lambda_{L,\min}=\frac{1}{R}$
• longest wavelength: transition $2\to 1$ $\frac{1}{\lambda_{L,\max}}=R\left(1-\frac{1}{4}\right)=\frac{3R}{4}$ So, $\lambda_{L,\max}=\frac{4}{3R}$

Thus, $\lambda \in \left[\frac{1}{R},\,\frac{4}{3R}\right]$

Balmer range

$\lambda \in \left[\frac{4}{R},\,\frac{36}{5R}\right]$

Since $\frac{4}{3R}<\frac{4}{R}$ there is a clear gap between the two ranges, so they do not overlap.

So D is correct.

6. Final answer

Correct statements are:

• A
• D