the ratio of the wavelengths of alpha line of lyman series in H atom and beta line of balmer series in he+ is

Answer

To find the ratio of the wavelengths of the alpha line of the Lyman series in the hydrogen atom (H) and the beta line of the Balmer series in the helium ion (He+), we can use the Rydberg formula for the wavelengths of spectral lines in hydrogen-like atoms.

### Step-by-Step Solution:

1. **Identify the transitions:**
- The alpha line of the Lyman series corresponds to the transition from n=2 to n=1 in hydrogen (H).
- The beta line of the Balmer series corresponds to the transition from n=4 to n=2 in helium ion (He+).

2. **Use the Rydberg formula:**
The Rydberg formula for the wavelength of emitted light during electron transitions in hydrogen-like atoms 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 (\( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and upper energy levels, respectively.

3. **Calculate the wavelength for the alpha line of Lyman series in H:**
- For hydrogen (H), \( Z = 1 \), \( n_1 = 1 \), \( n_2 = 2 \):
\[ \frac{1}{\lambda_H} = R \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]
Thus,
\[ \lambda_H = \frac{4}{3R} \]

4. **Calculate the wavelength for the beta line of Balmer series in He+:**
- For He+, \( Z = 2 \), \( n_1 = 2 \), \( n_2 = 4 \):
\[ \frac{1}{\lambda_{He^+}} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \cdot 4 \left( \frac{1}{4} - \frac{1}{16} \right) = R \cdot 4 \left( \frac{4 - 1}{16} \right) = R \cdot 4 \cdot \frac{3}{16} \]
Thus,
\[ \lambda_{He^+} = \frac{16}{12R} = \frac{4}{3R} \]

5. **Calculate the ratio of the wavelengths:**
\[ \frac{\lambda_H}{\lambda_{He^+}} = \frac{\frac{4}{3R}}{\frac{4}{3R}} = 1 \]

### Final Answer:
The ratio of the wavelengths of the alpha line of the Lyman series in H atom and the beta line of the Balmer series in He+ is **1**.