Question

Ratio of longest wavelength corresponding to Lyman and Balmer series in hydrogen spectrum is

• A. $\frac{7}{29}$
• B. $\frac{9}{31}$
• C. $\frac{5}{27}$
• D. $\frac{3}{23}$

Hint:

For the longest wavelength in a spectral series, the electron transition occurs from the next higher energy level (i.e., $n_2 = n_1+1$) to the base level ($n_1$). Pay close attention to the order requested for the ratio.

Answer 1

The Correct Option is A

Concept:
For hydrogen spectrum, wavelength of emitted line is given by Rydberg formula: \[ \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \] where:

• $R$ = Rydberg constant
• $n_1$ = lower energy level
• $n_2$ = higher energy level

For the longest wavelength in a series, energy gap must be minimum, so transition occurs from: \[ n_2=n_1+1 \]
Step 1: Longest wavelength of Lyman series
For Lyman series: \[ n_1=1 \] So longest wavelength transition: \[ n_2=2 \rightarrow n_1=1 \] Using formula: \[ \frac{1}{\lambda_L}=R\left(1-\frac{1}{4}\right) \] \[ \frac{1}{\lambda_L}=R\left(\frac{3}{4}\right) \] \[ \lambda_L=\frac{4}{3R} \]
Step 2: Longest wavelength of Balmer series
For Balmer series: \[ n_1=2 \] So longest wavelength transition: \[ n_2=3 \rightarrow n_1=2 \] \[ \frac{1}{\lambda_B}=R\left(\frac{1}{4}-\frac{1}{9}\right) \] \[ \frac{1}{\lambda_B}=R\left(\frac{5}{36}\right) \] \[ \lambda_B=\frac{36}{5R} \]
Step 3: Find ratio
\[ \frac{\lambda_L}{\lambda_B} = \frac{\frac{4}{3R{\frac{36}{5R \] \[ = \frac{4}{3R}\times\frac{5R}{36} \] \[ = \frac{20}{108} \] \[ = \frac{5}{27} \]
Step 4: Final Answer
The ratio of longest wavelength of Lyman series to Balmer series is: \[ \boxed{5:27} \] Quick Tip:
Longest wavelength in any series always comes from the nearest transition to lower level.