When 50 keV electrons are made incident on a target material, the wavelength of `K_(alpha)` X–ray line was found to be 0.5Å. When the accelerating potential is increased to 100 kV, then the wavelength of `K_(alpha)` –line from the same target will be

Options

A. 0.25 Å
B. 0.5 Å
C. 0.75 Å
D. 1.0 Å

Answer

To solve the problem, we need to understand the relationship between the accelerating potential of electrons and the wavelength of the emitted X-rays, specifically the K-alpha line.

### Step-by-Step Solution:

1. **Understanding the K-alpha X-ray Production**:
- K-alpha X-rays are produced when an electron from the L-shell (second shell) transitions to fill a vacancy in the K-shell (first shell) of an atom. The energy difference between these two shells corresponds to the energy of the emitted X-ray.

2. **Initial Conditions**:
- We are given that when 50 kV electrons are incident on the target, the wavelength of the K-alpha X-ray line is 0.5 Å.

3. **Energy of the Incident Electrons**:
- The energy of the incident electrons can be calculated using the formula:
\[ E = eV \]
where \(E\) is the energy in electron volts (eV), \(e\) is the charge of an electron, and \(V\) is the accelerating potential in volts (V).
- For 50 kV, the energy is:
\[ E_{50} = 50 \text{ keV} = 50,000 \text{ eV} \]

4. **Calculating the Wavelength**:
- The wavelength (\(\lambda\)) of the emitted X-ray can be calculated using the formula:
\[ \lambda = \frac{hc}{E} \]
where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \text{ Js}\)) and \(c\) is the speed of light (\(3 \times 10^8 \text{ m/s}\)).
- For the initial condition, we can verify that:
\[ \lambda_{50} = 0.5 \text{ Å} = 0.5 \times 10^{-10} \text{ m} \]

5. **Increasing the Accelerating Potential**:
- Now, when the accelerating potential is increased to 100 kV, the energy of the incident electrons becomes:
\[ E_{100} = 100 \text{ keV} = 100,000 \text{ eV} \]

6. **Effect on K-alpha Wavelength**:
- The K-alpha X-ray wavelength is primarily determined by the energy difference between the K and L shells, which does not change with the accelerating voltage. Therefore, even though the energy of the incident electrons has increased, the energy of the K-alpha transition remains the same.
- Thus, the wavelength of the K-alpha line will remain the same:
\[ \lambda_{100} = \lambda_{50} = 0.5 \text{ Å} \]

### Final Answer:
The wavelength of the K-alpha line from the same target when the accelerating potential is increased to 100 kV will still be **0.5 Å**.