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.

Options

A. minimum wavelength will be `62.1 nm`.
B. energy of the characteristic x-rays will be equal to oe less than `19.9 ke V`.
C. `L_(alpha)` X-ray may be emitted.
D. `L_(alpha)` X-ray will have energy `19.9 ke V`.

Explanation

To solve the problem step by step, we will analyze the given information and calculate the required values.

### Step 1: Understand the Given Information
- The voltage applied in the X-ray tube is \( V = 20 \, \text{kV} \).
- The energy required to remove an electron from the L shell is \( E_L = 19.9 \, \text{keV} \).

### Step 2: Calculate the Minimum Wavelength
The minimum wavelength of the X-rays emitted can be calculated using the formula:
\[ \lambda_{\text{min}} = \frac{hc}{E} \]
where:
- \( h \) is Planck's constant (\( 4.1357 \times 10^{-15} \, \text{eV s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( E \) is the energy in electron volts.

Since we are using \( E = 20 \, \text{kV} = 20,000 \, \text{eV} \), we can also use the simplified formula:
\[ \lambda_{\text{min}} = \frac{12400}{E} \, \text{(in nm)} \]
Substituting \( E = 20 \, \text{kV} = 20,000 \, \text{eV} \):
\[ \lambda_{\text{min}} = \frac{12400}{20} = 620 \, \text{nm} \]

### Step 3: Analyze the Characteristic X-ray Energy
The energy of characteristic X-rays is determined by the transition of electrons between energy levels. The energy of the emitted X-ray when an electron transitions from the M shell to the L shell (L-alpha transition) is equal to the energy difference between these two levels.

Since the energy required to remove an electron from the L shell is \( 19.9 \, \text{keV} \), the energy of the characteristic X-ray will be equal to or less than this value.

### Step 4: Evaluate the Options
- **Option A**: Minimum wavelength will be \( 62.1 \, \text{nm} \). (This is incorrect as we calculated \( 620 \, \text{nm} \)).
- **Option B**: The energy of characteristic X-ray will be equal to or less than \( 19.9 \, \text{keV} \). (This is correct).
- **Option C**: L-alpha X-ray may be emitted. (This is correct as transitions can occur).
- **Option D**: L-alpha X-ray will have energy \( 19.9 \, \text{keV} \). (This is correct as it can be equal to this value).

### Conclusion
The correct options are B, C, and D.