Choose the correct alternatives

Options

A. Ratio to maximum of minimum wavelength obtained in Balmer series of hydrogen spectrum is 1.8
B. For transition of electron from orbit with n=8 to orbit with n=3 , ultraviolet radiations are emitted
C. Lyman series corresponds to ultraviolet region of radiation
D. Spectrum of hydrogen is continuous

Answer

To solve the question, we need to evaluate each option provided regarding the hydrogen spectrum and determine which ones are correct.

### Step-by-Step Solution:

**Option A: The ratio of maximum to minimum wavelength obtained in the Balmer series of hydrogen spectrum is 1.8.**

1. The formula for the wavelength in the Balmer series is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right) \]
where \( R_H \) is the Rydberg constant, and \( n \) can take values 3, 4, 5, etc.

2. Calculate the minimum wavelength when \( n \) approaches infinity:
\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - 0 \right) = \frac{R_H}{4} \]
Thus, \( \lambda_1 = \frac{4}{R_H} \).

3. Calculate the maximum wavelength when \( n = 3 \):
\[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{9 - 4}{36} \right) = \frac{5R_H}{36} \]
Thus, \( \lambda_2 = \frac{36}{5R_H} \).

4. Now, find the ratio of maximum to minimum wavelengths:
\[ \frac{\lambda_{\text{max}}}{\lambda_{\text{min}}} = \frac{\lambda_2}{\lambda_1} = \frac{\frac{36}{5R_H}}{\frac{4}{R_H}} = \frac{36}{5} \cdot \frac{R_H}{4} = \frac{36}{20} = 1.8 \]

**Conclusion for Option A:** This option is correct.

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**Option B: For the transition of an electron from orbit with \( n = 8 \) to orbit with \( n = 3 \), ultraviolet radiation is emitted.**

1. The transition from a higher energy level to a lower one emits radiation. The formula for wavelength is:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{3^2} - \frac{1}{8^2} \right) \]
Calculate:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{9} - \frac{1}{64} \right) = R_H \left( \frac{64 - 9}{576} \right) = \frac{55R_H}{576} \]

2. The wavelength \( \lambda \) can be calculated as:
\[ \lambda = \frac{576}{55R_H} \]
Given that \( R_H \approx 911 \) Å, we can find \( \lambda \):
\[ \lambda \approx \frac{576 \times 911}{55} \approx 9540 \text{ Å} \]
This wavelength is in the infrared region, not ultraviolet.

**Conclusion for Option B:** This option is incorrect.

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**Option C: Lyman series corresponds to the ultraviolet region of radiation.**

1. The Lyman series involves transitions to \( n = 1 \) from higher levels (n = 2, 3, ...). All wavelengths in the Lyman series fall within the ultraviolet region.

**Conclusion for Option C:** This option is correct.

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**Option D: The spectrum of hydrogen is continuous.**

1. The spectrum of hydrogen is not continuous; it consists of discrete lines corresponding to specific transitions between energy levels.

**Conclusion for Option D:** This option is incorrect.

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### Final Answers:
- **Correct Options:** A, C
- **Incorrect Options:** B, D