Question

The shortest wavelength for Lyman series is $912\ \text{\AA}$. The longest wavelength in Paschen series is

• A. $1216\ \text{\AA}$
• B. $3646\ \text{\AA}$
• C. $18760\ \text{\AA}$
• D. $8208\ \text{\AA}$

Hint:

The shortest wavelength of any hydrogen spectral series is simply given by $\lambda_{\text{min}} = \frac{n_1^2}{R}$. For Lyman, $\lambda_L = \frac{1}{R} = 912\ \text{\AA}$. For the longest wavelength of a series, use the shortcut factor: $\lambda_{\text{max}} = \lambda_{\text{min}} \times \left[\frac{(n_1+1)^2}{(n_1+1)^2 - n_1^2}\right]$. For Paschen ($n_1=3$), this yields $\lambda_P = (912 \times 9) \times \frac{16}{7} = 18760\ \text{\AA}$.

Answer 1

The Correct Option is C

Step 1: Read the given data.
The shortest (series-limit) wavelength of the Lyman series is $912\,\text{\AA}$. We need the longest wavelength of the Paschen series.
Step 2: Use Rydberg's formula.
$\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$.
Step 3: Pin down the Rydberg constant from Lyman.
The Lyman series limit is $n_1 = 1$, $n_2 = \infty$, so $\dfrac{1}{\lambda_L} = R$, giving $\dfrac{1}{R} = 912\,\text{\AA}$.
Step 4: Set up the longest Paschen line.
Longest wavelength means smallest energy jump, i.e. $n_1 = 3$, $n_2 = 4$: $\dfrac{1}{\lambda_P} = R\left(\dfrac{1}{9} - \dfrac{1}{16}\right)$.
Step 5: Combine the fractions.
$\dfrac{1}{9} - \dfrac{1}{16} = \dfrac{16 - 9}{144} = \dfrac{7}{144}$, so $\lambda_P = \dfrac{144}{7R} = \dfrac{144}{7}\cdot\dfrac{1}{R}$.
Step 6: Substitute the value of $\frac{1}{R}$.
$\lambda_P = \dfrac{144}{7}\times 912 \approx 18760\,\text{\AA}$, which is option (3).
\[ \boxed{\lambda_P \approx 18760\ \text{\AA}} \]