The shortest wavelength produced in an X-ray tube operating at `0.5` million volt is

Options

A. dependent on the target element
B. about `2.5 xx 10^(-12) m`
C. double of the shortest wavelength produced at `1` milllion volt
D. dependent only on the target material

Explanation

To find the shortest wavelength produced in an X-ray tube operating at 0.5 million volts, we can use the formula for the minimum wavelength (λ_min) of X-rays:

\[ \lambda_{min} = \frac{hc}{eV} \]

where:
- \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) is the speed of light, approximately \( 3.00 \times 10^{8} \, \text{m/s} \)
- \( e \) is the charge of an electron, approximately \( 1.602 \times 10^{-19} \, \text{C} \)
- \( V \) is the accelerating potential in volts (0.5 million volts = \( 0.5 \times 10^6 \, \text{V} \))

### Step-by-Step Solution:

1. **Convert the potential to standard units**:
\[ V = 0.5 \, \text{million volts} = 0.5 \times 10^6 \, \text{V} = 5 \times 10^5 \, \text{V} \]

2. **Substitute the known values into the formula**:
\[ \lambda_{min} = \frac{(6.626 \times 10^{-34} \, \text{Js}) \times (3.00 \times 10^{8} \, \text{m/s})}{(1.602 \times 10^{-19} \, \text{C}) \times (5 \times 10^5 \, \text{V})} \]

3. **Calculate the numerator**:
\[ \text{Numerator} = 6.626 \times 10^{-34} \times 3.00 \times 10^{8} = 1.9878 \times 10^{-25} \, \text{Jm} \]

4. **Calculate the denominator**:
\[ \text{Denominator} = 1.602 \times 10^{-19} \times 5 \times 10^5 = 8.01 \times 10^{-14} \, \text{J} \]

5. **Divide the numerator by the denominator**:
\[ \lambda_{min} = \frac{1.9878 \times 10^{-25}}{8.01 \times 10^{-14}} \approx 2.48 \times 10^{-12} \, \text{m} \]

6. **Convert to picometers (pm)**:
\[ \lambda_{min} \approx 2.48 \times 10^{-12} \, \text{m} = 24.8 \, \text{pm} \]

### Conclusion:
The shortest wavelength produced in an X-ray tube operating at 0.5 million volts is approximately \( 2.48 \times 10^{-12} \, \text{m} \).