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

Options

A. `3/23`
B. `7/29`
C. `9/31`
D. `5/27`

Answer

To find the ratio of the longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum, we can follow these steps:

### Step 1: Understand the Series
The Lyman series corresponds to transitions where the final energy level (nf) is 1, while the Balmer series corresponds to transitions where nf is 2. For the longest wavelength in each series, we need to consider the transitions from the nearest higher energy levels.

### Step 2: Calculate the Longest Wavelength for Lyman Series
For the Lyman series, the longest wavelength corresponds to the transition from n = 2 to n = 1.

Using the formula for the wavelength in the hydrogen spectrum:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( n_f = 1 \) and \( n_i = 2 \).

Substituting the values:
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) \]
Thus,
\[ \lambda_1 = \frac{4}{3R} \]

### Step 3: Calculate the Longest Wavelength for Balmer Series
For the Balmer series, the longest wavelength corresponds to the transition from n = 3 to n = 2.

Using the same formula:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( n_f = 2 \) and \( n_i = 3 \).

Substituting the values:
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda_2} = R \left( \frac{9}{36} - \frac{4}{36} \right) = R \left( \frac{5}{36} \right) \]
Thus,
\[ \lambda_2 = \frac{36}{5R} \]

### Step 4: Calculate the Ratio of the Wavelengths
Now, we can find the ratio of the longest wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3R} \times \frac{5R}{36} = \frac{4 \times 5}{3 \times 36} = \frac{20}{108} = \frac{5}{27} \]

### Final Answer
The ratio of the longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is:
\[ \frac{5}{27} \]

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