In an X-ray tube the accelerating voltage is `20 kV`. Two target `A` and `B` are used one by one. For `'A'` the wavelength of the `K_(alpha)` lines is `62 "pm"`. For `'B'` the wavelength of the `L_(alpha)` line is `124 "pm"`. The energy of the `'B'` ion with vacancy in `'M'` shell is `5.5 KeV` higher than the atom of `B`. [ Take `hc=12400 eV Å`]

Options

A. Value of `lambda_(min)` is `0.62Å`
B. `A` will emitte `K_(alpha)`-photon.
C. `B` will emitte `L`-photns.
D. minimum wavelength (in `Å`) of the characterstics X-ray that will be emitted by `'B'` is `0.8Å`.

Explanation

To solve the problem step by step, we will analyze the information given and apply relevant formulas from atomic physics.

### Step 1: Calculate the minimum wavelength (λ_min)
The minimum wavelength (λ_min) in an X-ray tube can be calculated using the formula:

\[ \lambda_{\text{min}} = \frac{hc}{E} \]

Where:
- \( h \) is Planck's constant (given as \( hc = 12400 \, \text{eV} \cdot \text{Å} \))
- \( E \) is the energy of the accelerated electrons in eV.

Given that the accelerating voltage is \( 20 \, \text{kV} = 20000 \, \text{eV} \), we can substitute this into the equation.

\[ \lambda_{\text{min}} = \frac{12400 \, \text{eV} \cdot \text{Å}}{20000 \, \text{eV}} = 0.62 \, \text{Å} \]

### Step 2: Analyze the Kα line for target A
For target A, the wavelength of the Kα line is given as \( 62 \, \text{pm} \) (which is \( 0.62 \, \text{Å} \)). Since the minimum wavelength calculated is equal to the Kα wavelength, this means that the Kα line can be emitted.

### Step 3: Analyze the Lα line for target B
For target B, the wavelength of the Lα line is given as \( 124 \, \text{pm} \) (which is \( 1.24 \, \text{Å} \)). We need to determine if this line can be emitted.

### Step 4: Determine the energy required to remove an electron from the L shell
The energy of the B ion with a vacancy in the M shell is given as \( 5.5 \, \text{keV} \) higher than the atom of B. This means that the energy required to remove an electron from the L shell must be calculated.

Using the energy of the Lα line, we can find the energy required for the transition. The energy corresponding to the wavelength can be calculated by:

\[ E = \frac{hc}{\lambda} \]

Substituting for the Lα line:

\[ E_{L\alpha} = \frac{12400 \, \text{eV} \cdot \text{Å}}{1.24 \, \text{Å}} \approx 10000 \, \text{eV} = 10 \, \text{keV} \]

### Step 5: Check if the L shell electron can be removed
The total energy available from the accelerating voltage is \( 20 \, \text{keV} \). Since the energy required to remove the L shell electron is \( 10 \, \text{keV} \) and the energy available is \( 20 \, \text{keV} \), we can conclude that the L shell electron can indeed be removed.

### Step 6: Calculate the minimum wavelength for the emitted characteristic X-ray from B
Using the energy of the Lα transition calculated earlier, we can find the minimum wavelength:

\[ \lambda_{\text{min}} = \frac{hc}{E} \]

Substituting \( E = 10 \, \text{keV} \):

\[ \lambda_{\text{min}} = \frac{12400 \, \text{eV} \cdot \text{Å}}{10000 \, \text{eV}} = 1.24 \, \text{Å} \]

### Summary of Results
1. Kα line from target A can be emitted since \( \lambda_{\text{min}} = 0.62 \, \text{Å} \).
2. Lα line from target B can be emitted since the energy required is less than the available energy.
3. The minimum wavelength for the emitted characteristic X-ray from B is \( 1.24 \, \text{Å} \).