In an X-ray tube, the voltage applied is `20 kV`. The energy required to remove an electron from `L` shell is `19.9 ke V` In the X-rays emitted by the tube.

Answer

To find the minimum wavelength of the X-rays emitted from an X-ray tube with a voltage of 20 kV and an energy required to remove an electron from the L shell of 19.9 keV, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Relationship**:
The energy supplied to the electrons in the X-ray tube is given by the voltage applied. The energy (E) in electron volts (eV) can be calculated using the formula:
\[ E = e \cdot V \]
where \( e \) is the charge of an electron (approximately \( 1.6 \times 10^{-19} \) coulombs) and \( V \) is the voltage in volts.

2. **Calculate the Energy Supplied**:
Given that the voltage \( V = 20 \, \text{kV} = 20,000 \, \text{V} \):
\[ E = 1.6 \times 10^{-19} \, \text{C} \times 20,000 \, \text{V} = 3.2 \times 10^{-15} \, \text{J} \]
To convert this energy into electron volts:
\[ E = 20 \, \text{keV} = 20,000 \, \text{eV} \]

3. **Determine the Energy of the Photon**:
The energy required to remove an electron from the L shell is given as \( 19.9 \, \text{keV} \). This energy is the minimum energy of the emitted X-ray photon when the electron transitions from infinity back to the L shell.

4. **Use the Energy-Wavelength Relationship**:
The energy of a photon can also be expressed in terms of its wavelength (\( \lambda \)) using the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)) and \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).

5. **Rearranging for Wavelength**:
Rearranging the equation to solve for \( \lambda \):
\[ \lambda = \frac{hc}{E} \]

6. **Convert Energy to Joules**:
Convert \( 19.9 \, \text{keV} \) to joules:
\[ E = 19.9 \times 10^3 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.184 \times 10^{-15} \, \text{J} \]

7. **Calculate the Minimum Wavelength**:
Substitute the values into the wavelength formula:
\[ \lambda = \frac{(6.63 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{3.184 \times 10^{-15} \, \text{J}} \]
\[ \lambda \approx 6.25 \times 10^{-11} \, \text{m} = 62.5 \, \text{pm} \]

### Final Result:
The minimum wavelength of the X-rays emitted by the tube is approximately \( 62.5 \, \text{pm} \) (picometers).