Last line of Lyman series for `H-` atom has wavelength `lambda_(1) A,2^(nd)` line of Balmer series has wavelength `lambda_(2)A` then

Options

A. `(16)/(lambda_(1))=(9)/(lambda_(2))`
B. `(16)/(lambda_(2))=(3)/(lambda_(1))`
C. `(4)/(lambda_(1))=(1)/(lambda_(2))`
D. `(16)/(lambda_(1))=(3)/(lambda_(2))`

Explanation

To solve the problem, we need to find the ratio of the wavelengths of the last line of the Lyman series (λ₁) and the second line of the Balmer series (λ₂) for the hydrogen atom.

### Step 1: Understand the Series
The Lyman series corresponds to transitions where the electron falls to the n=1 level, while the Balmer series corresponds to transitions where the electron falls to the n=2 level.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of emitted light during electron transitions is given by:
\[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 3: Calculate λ₁ for the Lyman Series
For the last line of the Lyman series, the transition is from \( n = \infty \) to \( n = 1 \):
\[ \frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \left( 1 - 0 \right) = R \]
Thus, we have:
\[ \lambda_1 = \frac{1}{R} \]

### Step 4: Calculate λ₂ for the Balmer Series
For the second line of the Balmer series, the transition is from \( n = 4 \) to \( n = 2 \):
\[ \frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) \]
Calculating the right side:
\[ \frac{1}{4} - \frac{1}{16} = \frac{4}{16} - \frac{1}{16} = \frac{3}{16} \]
So, we have:
\[ \frac{1}{\lambda_2} = R \cdot \frac{3}{16} \]
Thus, we can express \( \lambda_2 \) as:
\[ \lambda_2 = \frac{16}{3R} \]

### Step 5: Find the Ratio of λ₁ to λ₂
Now we can find the ratio \( \frac{\lambda_1}{\lambda_2} \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{1}{R}}{\frac{16}{3R}} = \frac{1}{R} \cdot \frac{3R}{16} = \frac{3}{16} \]

### Final Answer
The ratio of the wavelengths is:
\[ \frac{\lambda_1}{\lambda_2} = \frac{3}{16} \]