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: Understanding the Question:
We are given the minimum wavelength limit of the Lyman spectral series for a hydrogen-like atom. We need to determine the maximum wavelength limit belonging to the Paschen series using Rydberg's formula.

Step 2: Key Formula or Approach:
Rydberg's formula for the wave number of spectral lines is:
$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ where $R$ is the Rydberg constant. 1. The shortest wavelength in the Lyman series ($\lambda_L$) occurs when an electron transitions from $n_2 = \infty$ to $n_1 = 1$.
2. The longest wavelength in the Paschen series ($\lambda_P$) corresponds to the minimum energy jump within that series, transitioning from the adjacent line $n_2 = 4$ to $n_1 = 3$.

Step 3: Detailed Explanation:
Let's find the mathematical expression for the shortest Lyman wavelength:
$$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \implies \lambda_L = \frac{1}{R} = 912\ \text{\AA}$$ Next, let's establish the equation for the longest Paschen wavelength:
$$\frac{1}{\lambda_P} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right)$$ Find a common denominator to subtract the fractions:
$$\frac{1}{\lambda_P} = R \left( \frac{16 - 9}{144} \right) = \frac{7R}{144} \implies \lambda_P = \frac{144}{7R}$$ Substitute the known value of $\frac{1}{R} = \lambda_L = 912\ \text{\AA}$ directly into this final expression:
$$\lambda_P = \frac{144}{7} \times 912\ \text{\AA}$$ $$\lambda_P = \frac{131328}{7}\ \text{\AA} \approx 18761.14\ \text{\AA}$$ Rounding to the nearest whole integer matches the value given in option (C).

Step 4: Final Answer:
The longest wavelength in the Paschen series is $18760\ \text{\AA}$, matching option (C).