In an X - ray tube , electrons emitted from a filament (cathode) carrying current hit a target (anode) at a distance d form the cathode. The target is kept at a potential V higher than the cathode resulting in emission of continuous of and charateristic X - rays. It the filament current I is decreased to `(I)/(2)` , the potential difference V is increaed to 2v, and the separation distance d is reduuced to `(d)/(2)` ,then

Options

A. The cut off wavelenghth will reduced to half, and the wavelenghts of the characteristi c X - rays wil reminan the same
B. the cut - off wavelength as well as the wavelenght of the characteristic X - rays will remain the same
C. the cut -off wavelenght will reduced to half , and the intensities of all the X - rays wil decrease
D. the cut - off wavelength will become two time larger, and the intensity of all the X -rays wil decrease

Explanation

To solve the problem regarding the changes in an X-ray tube when certain parameters are altered, we will analyze the effects of these changes step-by-step.

### Step 1: Understanding the Initial Conditions
In an X-ray tube, electrons are emitted from a cathode (filament) and accelerated towards an anode (target) due to a potential difference \( V \). The kinetic energy gained by the electrons is converted into X-rays upon collision with the anode.

### Step 2: Determine the Cut-off Wavelength
The cut-off wavelength \( \lambda_{cut-off} \) is related to the maximum energy of the electrons, which can be expressed as:
\[ E_{max} = eV \]
where \( e \) is the charge of the electron and \( V \) is the potential difference. The energy of the X-ray photon is also given by:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light. Setting these equal gives:
\[ \lambda_{cut-off} = \frac{hc}{eV} \]
This shows that the cut-off wavelength is inversely proportional to the potential \( V \).

### Step 3: Analyze the Changes
1. **Filament Current Decrease**: The filament current \( I \) is decreased to \( \frac{I}{2} \). This affects the number of electrons emitted but does not directly affect the cut-off wavelength.

2. **Potential Increase**: The potential \( V \) is increased to \( 2V \). Since the cut-off wavelength is inversely proportional to the potential, we have:
\[ \lambda_{cut-off}' = \frac{hc}{e(2V)} = \frac{1}{2} \cdot \frac{hc}{eV} = \frac{1}{2} \lambda_{cut-off} \]
Thus, the cut-off wavelength is reduced to half.

3. **Distance Decrease**: The separation distance \( d \) is reduced to \( \frac{d}{2} \). This does not directly affect the cut-off wavelength or the energy of the emitted X-rays.

### Step 4: Characteristic X-ray Wavelength
The wavelength of characteristic X-rays depends on the atomic number \( Z \) of the target material. Since the material of the anode is unchanged, the characteristic X-ray wavelength remains the same.

### Step 5: Intensity of X-rays
The intensity of the emitted X-rays is proportional to the number of electrons hitting the anode. Since the filament current is decreased, the number of emitted electrons will also decrease, leading to a decrease in the intensity of the X-rays.

### Conclusion
Based on the analysis:
- The cut-off wavelength is reduced to half.
- The wavelength of characteristic X-rays remains the same.
- The intensity of all X-rays decreases due to the reduced filament current.

### Final Answer
The correct options are:
1. The cut-off wavelength will reduce to half and the wavelength of the characteristic X-ray will remain the same.
3. The cut-off wavelength will reduce to half and the intensity of all the X-rays will decrease.