Assertion :The spectral series 'Blamer' of the of the hydrogen atom lies in the visible region of the electromagnetic spetrum
Reason :Wavelength of light in the visible region lies in the range of 400 nm to 700 nm

Options

A. If both Assertion and Reason ar true and Reason is the correct explanation of Assertion.
B. If both Assertion and Reason are true but Reason is not correct explanation of Assertion
C. If Assertion is true by Reason is false.
D. If Assertion is false but Reason is true.

Answer

To solve the question, we need to analyze both the assertion and the reason provided.

### Step 1: Understand the Assertion
The assertion states that the Balmer series of the hydrogen atom lies in the visible region of the electromagnetic spectrum. The Balmer series corresponds to electronic transitions where an electron falls from a higher energy level (n ≥ 3) to the n = 2 energy level.

### Step 2: Determine the Wavelengths of the Balmer Series
To determine if the Balmer series lies in the visible region, we need to find the wavelengths of the spectral lines in this series. The wavelengths can be calculated using the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, m^{-1} \)), \( n_1 = 2 \) for the Balmer series, and \( n_2 \) can be 3, 4, 5, etc.

### Step 3: Calculate the Wavelengths
1. For \( n_2 = 3 \):
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{5}{36} \right) \]
Calculate \( \lambda \) and find it is approximately 656 nm.

2. For \( n_2 = 4 \):
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \left( \frac{3}{16} \right) \]
Calculate \( \lambda \) and find it is approximately 486 nm.

3. For \( n_2 = 5 \):
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{25} \right) = R_H \left( \frac{21}{100} \right) \]
Calculate \( \lambda \) and find it is approximately 434 nm.

4. For \( n_2 = 6 \):
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{6^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{36} \right) = R_H \left( \frac{8}{36} \right) \]
Calculate \( \lambda \) and find it is approximately 410 nm.

### Step 4: Compare with the Visible Spectrum
The visible spectrum ranges from approximately 400 nm to 700 nm. The wavelengths calculated for the Balmer series (410 nm to 656 nm) indeed fall within this range.

### Conclusion
Both the assertion and the reason are true:
- The Balmer series does lie in the visible region of the electromagnetic spectrum.
- The reason correctly states that the wavelength of light in the visible region is between 400 nm and 700 nm.

Thus, the answer is that both the assertion and reason are true, but the reason does not explain the assertion.

### Final Answer
Assertion: True
Reason: True
Conclusion: Reason does not explain the assertion.