The minimum wavelength of the X- rays produced at accelerationg potential V is `lambda`. If the accelerating potential is changed to 2V, then the minimum wavelength would become

Options

A. `4lambda`
B. `2lambda`
C. `lambda/2`
D. `lambda/4`

Answer

To solve the problem, we need to understand the relationship between the accelerating potential (V) and the minimum wavelength (λ) of the X-rays produced.

### Step-by-Step Solution:

1. **Understanding the relationship**: The energy of the electrons accelerated through a potential difference V is given by the equation:
\[ E = eV \]
where \( e \) is the charge of the electron.

2. **Energy and wavelength relationship**: The energy of the emitted X-ray photons is related to their wavelength by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

3. **Setting up the equation**: When the electrons collide with the target material, the energy of the electrons is converted into X-ray energy. Therefore, we can equate the two expressions for energy:
\[ eV = \frac{hc}{\lambda} \]

4. **Finding the minimum wavelength**: Rearranging the equation gives us:
\[ \lambda = \frac{hc}{eV} \]
This shows that the wavelength is inversely proportional to the accelerating potential \( V \).

5. **Doubling the potential**: If we change the accelerating potential from \( V \) to \( 2V \), we can express the new minimum wavelength \( \lambda' \) as:
\[ \lambda' = \frac{hc}{e(2V)} = \frac{hc}{2eV} \]
This simplifies to:
\[ \lambda' = \frac{1}{2} \cdot \frac{hc}{eV} = \frac{\lambda}{2} \]

6. **Conclusion**: Therefore, when the accelerating potential is changed to \( 2V \), the new minimum wavelength becomes:
\[ \lambda' = \frac{\lambda}{2} \]

### Final Answer:
The minimum wavelength would become \( \frac{\lambda}{2} \).

---