Question

In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

• A. $\frac{27}{5}$
• B. $\frac{5}{27}$
• C. $\frac{4}{9}$
• D. $\frac{9}{4}$

Answer 1

The Correct Option is B

To find the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series in the hydrogen spectrum, we need to understand the transitions in these series:

1. The Lyman series consists of electron transitions from higher energy levels (n ≥ 2) to the first energy level (n = 1).
2. The formula for the wavelength of emitted light in the Lyman series is given by the Rydberg formula: \[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{n^2} \right) \] where R is the Rydberg constant, and n is the principal quantum number of the higher energy level.
3. The longest wavelength in the Lyman series corresponds to the transition from n = 2 to n = 1 , or \frac{1}{\lambda_L} = R \left( 1 - \frac{1}{2^2} \right) = R \cdot \frac{3}{4} \right) \] .
4. The Balmer series consists of transitions from higher energy levels (n ≥ 3) to the second energy level (n = 2).
5. The formula for the wavelength in the Balmer series is: \[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \]
6. The longest wavelength in the Balmer series corresponds to the transition from n = 3 to n = 2 , or \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \cdot \frac{5}{36} \right) \] .

Now, the ratio of the longest wavelength in the Lyman series ( \lambda_L ) to the longest wavelength in the Balmer series ( \lambda_B ) is:

\[ \text{Ratio} = \frac{\lambda_L}{\lambda_B} = \frac{\frac{4}{3R}}{\frac{36}{5R}} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27} \]

Thus, the correct answer is \frac{5}{27} .