Calculate the wavelength of `K_beta` line (in `Å` ) emitted by a hydrogen-like element for which the wavelength of the characteristic X-ray `K_(alpha)` is known to be 0.32 `Å`

Explanation

To calculate the wavelength of the `K_beta` line emitted by a hydrogen-like element, we can use the relationship between the wavelengths of the `K_alpha` and `K_beta` lines. The steps are as follows:

### Step-by-Step Solution:

1. **Identify the given information:**
- Wavelength of `K_alpha`, \( \lambda_{K_{\alpha}} = 0.32 \, \text{Å} \)

2. **Understand the transitions:**
- For `K_alpha`, the transition is from \( n_2 = 2 \) to \( n_1 = 1 \).
- For `K_beta`, the transition is from \( n_2 = 3 \) to \( n_1 = 1 \).

3. **Use the formula for wavelength:**
The formula for the wavelength of X-rays emitted during electronic 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, and \( Z \) is the atomic number of the element.

4. **Calculate for `K_alpha`:**
For `K_alpha`:
\[ \frac{1}{\lambda_{K_{\alpha}}} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
\[ \frac{1}{\lambda_{K_{\alpha}}} = RZ^2 \left( 1 - \frac{1}{4} \right) = RZ^2 \left( \frac{3}{4} \right) \]

5. **Calculate for `K_beta`:**
For `K_beta`:
\[ \frac{1}{\lambda_{K_{\beta}}} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) \]
\[ \frac{1}{\lambda_{K_{\beta}}} = RZ^2 \left( 1 - \frac{1}{9} \right) = RZ^2 \left( \frac{8}{9} \right) \]

6. **Relate the two wavelengths:**
Since both expressions involve \( RZ^2 \), we can relate them:
\[ \frac{1}{\lambda_{K_{\beta}}} = \frac{8}{9} \cdot \frac{1}{\lambda_{K_{\alpha}}} \cdot \frac{3}{4} \]
Rearranging gives:
\[ \lambda_{K_{\beta}} = \lambda_{K_{\alpha}} \cdot \frac{9}{8} \cdot \frac{4}{3} \]

7. **Substitute the value of \( \lambda_{K_{\alpha}} \):**
\[ \lambda_{K_{\beta}} = 0.32 \cdot \frac{9}{8} \cdot \frac{4}{3} \]

8. **Calculate \( \lambda_{K_{\beta}} \):**
\[ \lambda_{K_{\beta}} = 0.32 \cdot \frac{36}{24} = 0.32 \cdot 1.5 = 0.48 \, \text{Å} \]

### Final Answer:
The wavelength of the `K_beta` line is \( \lambda_{K_{\beta}} = 0.48 \, \text{Å} \).