A beam of electron accelerated by a large potential difference `V` is mode of strike a metal target of produce X-ray . For what value of `V` , will the resulting X-ray have the lower minimum wavelength?

Explanation

To solve the problem of finding the potential difference \( V \) that results in the minimum wavelength of X-rays produced when a beam of electrons strikes a metal target, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship Between Energy and Wavelength**:
The energy \( E \) of a photon is related to its wavelength \( \lambda \) by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

2. **Relating Energy to Potential Difference**:
When electrons are accelerated through a potential difference \( V \), they gain kinetic energy equal to the work done on them by the electric field:
\[ E = eV \]
where \( e \) is the charge of an electron.

3. **Combining the Equations**:
By equating the two expressions for energy, we have:
\[ eV = \frac{hc}{\lambda} \]
Rearranging this gives us the wavelength in terms of the potential difference:
\[ \lambda = \frac{hc}{eV} \]

4. **Finding the Minimum Wavelength**:
From the equation \( \lambda = \frac{hc}{eV} \), we can see that the wavelength \( \lambda \) is inversely proportional to the potential difference \( V \). This means that to minimize \( \lambda \), we need to maximize \( V \).

5. **Choosing the Maximum Potential Difference**:
Given multiple options for the potential difference \( V \), we need to select the largest value to achieve the minimum wavelength.

6. **Identifying the Correct Option**:
If the options provided are:
- Option 1: 20 kV
- Option 2: 30 kV
- Option 3: 35 kV
- Option 4: 40 kV

The largest potential difference is 40 kV.

### Conclusion:
Thus, the value of \( V \) that will result in the minimum wavelength of the X-rays produced is:
\[ \boxed{40 \text{ kV}} \]