The cut-off wavelength of an X-ray depends on

Options

A. accelerating voltage only
B. both the accelerating voltage and material of the target
C. both the filament current and accelerating voltage
D. filament current only

Answer

To solve the question regarding the cut-off wavelength of X-rays, we will follow these steps:

### Step 1: Understand the Concept of Cut-off Wavelength
The cut-off wavelength (\( \lambda_{\text{cut-off}} \)) of X-rays is the maximum wavelength of X-rays that can be produced when high-energy electrons strike a target material. It is related to the energy of the X-rays.

### Step 2: Relate Energy to Wavelength
The energy (\( E \)) of an X-ray photon can be expressed using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the X-ray.

### Step 3: Relate Energy to Accelerating Voltage
When electrons are accelerated through a potential difference (accelerating voltage, \( V \)), they gain kinetic energy, which can be expressed as:
\[ E = eV \]
where:
- \( e \) is the charge of the electron,
- \( V \) is the accelerating voltage.

### Step 4: Equate the Two Expressions for Energy
Setting the two expressions for energy equal gives:
\[ eV = \frac{hc}{\lambda_{\text{cut-off}}} \]

### Step 5: Solve for Cut-off Wavelength
Rearranging the equation to solve for the cut-off wavelength yields:
\[ \lambda_{\text{cut-off}} = \frac{hc}{eV} \]

### Step 6: Analyze the Relationship
From the equation \( \lambda_{\text{cut-off}} = \frac{hc}{eV} \), we see that:
- \( h \) and \( c \) are constants,
- \( e \) is a constant (the charge of the electron),
- The cut-off wavelength is inversely proportional to the accelerating voltage \( V \).

### Step 7: Conclusion
Thus, the cut-off wavelength of X-rays depends only on the accelerating voltage and not on the material of the target or filament current. Therefore, the correct answer is:
**Option A: Accelerating voltage only.**