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. The key concept here is that the energy of the X-rays is related to the accelerating potential.

### Step-by-Step Solution:

1. **Understanding the Energy Relationship**:
The energy (E) of the X-rays can be expressed using the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

2. **Energy Gained from Accelerating Potential**:
When a charge \( Q \) is accelerated through a potential \( V \), it gains energy given by:
\[ E = QV \]

3. **Setting the Energies Equal**:
Since the energy gained by the charge is equal to the energy of the X-rays, we can set the two expressions for energy equal to each other:
\[ \frac{hc}{\lambda} = QV \]

4. **Finding the Minimum Wavelength**:
Rearranging the equation gives us the expression for the minimum wavelength:
\[ \lambda = \frac{hc}{QV} \]

5. **Changing the Accelerating Potential**:
Now, if we change the accelerating potential from \( V \) to \( 2V \), we need to find the new minimum wavelength \( \lambda_2 \):
\[ \lambda_2 = \frac{hc}{Q(2V)} = \frac{hc}{2QV} \]

6. **Relating the New Wavelength to the Old Wavelength**:
We can express \( \lambda_2 \) in terms of \( \lambda \):
\[ \lambda_2 = \frac{1}{2} \left(\frac{hc}{QV}\right) = \frac{\lambda}{2} \]

7. **Final Result**:
Therefore, when the accelerating potential is changed to \( 2V \), the minimum wavelength becomes:
\[ \lambda_2 = \frac{\lambda}{2} \]

### Conclusion:
The minimum wavelength of the X-rays produced at an accelerating potential of \( 2V \) is \( \frac{\lambda}{2} \).