In X-ray tube , when the accelerating voltage `V` is halved, the difference between the wavelength of `K_(alpha)` line and minimum wavelength of continuous X-ray spectrum

Options

A. remains constant
B. becomes more than two times
C. becomes half
D. becomes less than two times

Answer

To solve the problem, we need to analyze the relationship between the accelerating voltage in an X-ray tube and the wavelengths of the K-alpha line and the minimum wavelength of the continuous X-ray spectrum.

### Step-by-Step Solution:

1. **Understanding the Wavelengths**:
- The minimum wavelength (\( \lambda_{\text{min}} \)) of the continuous X-ray spectrum is inversely proportional to the accelerating voltage (\( V \)). The relationship can be expressed as:
\[ \lambda_{\text{min}} = \frac{hc}{eV} \]
- Here, \( h \) is Planck's constant, \( c \) is the speed of light, and \( e \) is the charge of the electron.

2. **Effect of Halving the Voltage**:
- When the accelerating voltage \( V \) is halved (\( V' = \frac{V}{2} \)), the new minimum wavelength (\( \lambda'_{\text{min}} \)) becomes:
\[ \lambda'_{\text{min}} = \frac{hc}{e \cdot \frac{V}{2}} = 2 \cdot \frac{hc}{eV} = 2 \lambda_{\text{min}} \]
- Thus, the new minimum wavelength is twice the original minimum wavelength.

3. **K-alpha Line Wavelength**:
- The wavelength of the K-alpha line (\( \lambda_{K\alpha} \)) remains constant regardless of the accelerating voltage.

4. **Calculating the Difference in Wavelengths**:
- The difference in wavelengths before halving the voltage is given by:
\[ \Delta \lambda = \lambda_{K\alpha} - \lambda_{\text{min}} \]
- After halving the voltage, the new difference in wavelengths (\( \Delta \lambda' \)) becomes:
\[ \Delta \lambda' = \lambda_{K\alpha} - \lambda'_{\text{min}} = \lambda_{K\alpha} - 2 \lambda_{\text{min}} \]

5. **Relating the New Difference to the Old Difference**:
- We can express the new difference in terms of the old difference:
\[ \Delta \lambda' = \lambda_{K\alpha} - 2 \lambda_{\text{min}} = (\lambda_{K\alpha} - \lambda_{\text{min}}) - \lambda_{\text{min}} = \Delta \lambda - \lambda_{\text{min}} \]

6. **Conclusion**:
- Since \( \Delta \lambda' \) is equal to \( \Delta \lambda - \lambda_{\text{min}} \), it is clear that \( \Delta \lambda' \) is less than \( 2 \Delta \lambda \) because \( \lambda_{\text{min}} \) is a positive quantity.
- Therefore, the final answer is that the difference between the wavelength of the K-alpha line and the minimum wavelength of the continuous X-ray spectrum becomes **less than 2 times** the original difference.

### Final Answer:
The difference between the wavelength of the K-alpha line and the minimum wavelength of the continuous X-ray spectrum becomes **less than 2 times**.