A X-ray tube operates at an accelerating potntial of 20 kV. Which of the following wavelength will be absent in the continuous spectrum of X-rays ?

Options

A. `12` pm
B. `75` pm
C. `65` pm
D. `95` pm

Answer

To solve the problem of determining which wavelength will be absent in the continuous spectrum of X-rays produced by an X-ray tube operating at an accelerating potential of 20 kV, we can follow these steps:

### Step 1: Understand the relationship between accelerating potential and minimum wavelength
The minimum wavelength (\( \lambda_{min} \)) of X-rays produced in an X-ray tube can be calculated using the formula:
\[ \lambda_{min} = \frac{hc}{eV} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( c \) is the speed of light (\( 3 \times 10^{8} \, \text{m/s} \))
- \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \))
- \( V \) is the accelerating potential (in volts)

### Step 2: Substitute the values into the formula
Given that the accelerating potential \( V = 20 \, \text{kV} = 20 \times 10^{3} \, \text{V} \), we can substitute the values into the formula:
\[ \lambda_{min} = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^{8} \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C})(20 \times 10^{3} \, \text{V})} \]

### Step 3: Calculate the minimum wavelength
Calculating the denominator:
\[ eV = (1.6 \times 10^{-19} \, \text{C})(20 \times 10^{3} \, \text{V}) = 3.2 \times 10^{-15} \, \text{J} \]

Now substituting back into the equation:
\[ \lambda_{min} = \frac{(6.626 \times 10^{-34})(3 \times 10^{8})}{3.2 \times 10^{-15}} \]

Calculating the numerator:
\[ 6.626 \times 10^{-34} \times 3 \times 10^{8} = 1.9878 \times 10^{-25} \, \text{Js m} \]

Now calculating \( \lambda_{min} \):
\[ \lambda_{min} = \frac{1.9878 \times 10^{-25}}{3.2 \times 10^{-15}} = 6.20 \times 10^{-11} \, \text{m} \]

### Step 4: Convert to picometers
Since \( 1 \, \text{pm} = 10^{-12} \, \text{m} \):
\[ \lambda_{min} = 6.20 \times 10^{-11} \, \text{m} = 62 \, \text{pm} \]

### Step 5: Determine which wavelengths are present or absent
The continuous spectrum of X-rays will include wavelengths greater than \( \lambda_{min} \). Therefore, wavelengths greater than 62 pm will be present, while those less than 62 pm will be absent.

Given the options:
- 12 pm (absent)
- 75 pm (present)
- 65 pm (present)
- 95 pm (present)

### Conclusion
The wavelength that will be absent in the continuous spectrum of X-rays is **12 pm**.

---