An X-ray tube is operated at `6.6 kV`. In the continuous spectrum of the emitted X-rays, which of the following frequency will be missing?

Options

A. wavelengths `0.01 nm` and `0.02 nm` will both be present
B. wavelength `0.01 nm` and `0.2 nm` will both be absent
C. wavelength `0.01 nm` will be present but wavelength `0.02 nm` will be absent
D. wavelength `0.01 nm` will be absent but wavelength `0.2 nm` will be present

Answer

To solve the problem of determining which frequency will be missing in the continuous spectrum of X-rays emitted from an X-ray tube operated at 6.6 kV, we can follow these steps:

### Step 1: Understand the relationship between energy, wavelength, and frequency
The energy \( E \) of a photon is related to its wavelength \( \lambda \) by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength.

### Step 2: Relate energy to the potential difference
The energy of the electrons accelerated through a potential difference \( V \) is given by:
\[ E = eV \]
where:
- \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \)),
- \( V \) is the potential difference (in volts).

### Step 3: Calculate the maximum energy of the emitted X-rays
Given that the X-ray tube is operated at \( 6.6 \, \text{kV} \), we convert this to volts:
\[ V = 6.6 \times 10^3 \, \text{V} \]
Now, we can calculate the maximum energy:
\[ E = eV = (1.6 \times 10^{-19} \, \text{C})(6.6 \times 10^3 \, \text{V}) = 1.056 \times 10^{-15} \, \text{J} \]

### Step 4: Calculate the minimum wavelength
Using the energy-wavelength relationship:
\[ \lambda = \frac{hc}{E} \]
Substituting the values:
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{1.056 \times 10^{-15} \, \text{J}} \approx 1.88 \times 10^{-11} \, \text{m} = 0.0188 \, \text{nm} \]

### Step 5: Determine the missing frequency
The frequency \( f \) can be calculated using the relationship:
\[ f = \frac{c}{\lambda} \]
Substituting the value of \( \lambda \):
\[ f = \frac{3 \times 10^8 \, \text{m/s}}{1.88 \times 10^{-11} \, \text{m}} \approx 1.60 \times 10^{19} \, \text{Hz} \]

### Step 6: Identify the missing frequency
In the continuous spectrum of emitted X-rays, the frequency corresponding to the minimum wavelength (maximum energy) will be missing. Therefore, the frequency \( f \approx 1.60 \times 10^{19} \, \text{Hz} \) will not be present in the spectrum.

### Conclusion
The frequency that will be missing in the continuous spectrum of the emitted X-rays is approximately \( 1.60 \times 10^{19} \, \text{Hz} \).