In a Coolidege tube experiment, the minimum wavelength of the continuouse X-ray spectrum is equal to `66.3` om . Then

Options

A. electrons accelerate through a potential difference of `12.75 kV` in the Coolidge tube
B. electrons accelerate through a potential difference of `18.75 kV` in the Coolidge tube
C. de-Broglie wavelength of the electrons reaching the anticathode
is of the order of `10 "pm"`
D. de-Broglie wavelength of the electrons reaching the anticathode is `0.01 Å`

Explanation

To solve the problem regarding the Coolidge tube experiment and the minimum wavelength of the continuous X-ray spectrum, follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The minimum wavelength (λ₀) of the continuous X-ray spectrum is given as 66.3 picometers (pm).
- Convert this wavelength into meters for calculations:
\[ λ₀ = 66.3 \, \text{pm} = 66.3 \times 10^{-12} \, \text{m} \]

2. **Use the Formula for Minimum Wavelength:**
- The formula relating minimum wavelength to the potential difference (Vₐ) in a Coolidge tube is:
\[ λ₀ = \frac{hc}{eVₐ} \]
- Rearranging the formula to find the potential difference:
\[ Vₐ = \frac{hc}{eλ₀} \]

3. **Substitute the Constants:**
- Use the following constants:
- Planck’s constant (h) = \(6.626 \times 10^{-34} \, \text{Js}\)
- Speed of light (c) = \(3.00 \times 10^{8} \, \text{m/s}\)
- Charge of an electron (e) = \(1.602 \times 10^{-19} \, \text{C}\)

4. **Calculate the Potential Difference:**
- Substitute the values into the equation:
\[ Vₐ = \frac{(6.626 \times 10^{-34} \, \text{Js})(3.00 \times 10^{8} \, \text{m/s})}{(1.602 \times 10^{-19} \, \text{C})(66.3 \times 10^{-12} \, \text{m})} \]
- Calculate the numerator:
\[ \text{Numerator} = 6.626 \times 10^{-34} \times 3.00 \times 10^{8} = 1.9878 \times 10^{-25} \, \text{Jm} \]
- Calculate the denominator:
\[ \text{Denominator} = 1.602 \times 10^{-19} \times 66.3 \times 10^{-12} = 1.063626 \times 10^{-30} \, \text{C m} \]
- Now, calculate \(Vₐ\):
\[ Vₐ = \frac{1.9878 \times 10^{-25}}{1.063626 \times 10^{-30}} \approx 18.75 \, \text{kV} \]

5. **Conclusion:**
- The calculated potential difference \(Vₐ\) is approximately 18.75 kV, which matches the second option provided in the question.