The ratio of wavelength of the lest line of Balmer series and the last line Lyman series is:

Options

A. 2
B. 1
C. 4
D. 0.5

Explanation

To find the ratio of the wavelength of the last line of the Balmer series to the last line of the Lyman series, we can use the Rydberg formula for hydrogen spectral lines. The formula is given by:

\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \( \lambda \) is the wavelength,
- \( R \) 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 1: Calculate the wavelength for the last line of the Balmer series

For the Balmer series, the transition occurs from \( n_2 = \infty \) to \( n_1 = 2 \).

Using the Rydberg formula:

\[ \frac{1}{\lambda_B} = R \cdot Z^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]

Since \( \frac{1}{\infty^2} = 0 \), we have:

\[ \frac{1}{\lambda_B} = R \cdot 1^2 \left( \frac{1}{4} \right) = \frac{R}{4} \]

Thus,

\[ \lambda_B = \frac{4}{R} \]

### Step 2: Calculate the wavelength for the last line of the Lyman series

For the Lyman series, the transition occurs from \( n_2 = \infty \) to \( n_1 = 1 \).

Using the Rydberg formula:

\[ \frac{1}{\lambda_L} = R \cdot Z^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]

Again, since \( \frac{1}{\infty^2} = 0 \), we have:

\[ \frac{1}{\lambda_L} = R \cdot 1^2 \left( 1 \right) = R \]

Thus,

\[ \lambda_L = \frac{1}{R} \]

### Step 3: Calculate the ratio of the wavelengths

Now, we need to find the ratio \( \frac{\lambda_B}{\lambda_L} \):

\[ \frac{\lambda_B}{\lambda_L} = \frac{\frac{4}{R}}{\frac{1}{R}} = 4 \]

### Conclusion

The ratio of the wavelength of the last line of the Balmer series to the last line of the Lyman series is:

\[ \frac{\lambda_B}{\lambda_L} = 4 \]

Thus, the answer is **4**.