If `V` be the accelerating voltage, then the maximum frequency of continuous X-rays is given by

Options

A. `(eh)/(V)`
B. `(hV)/(e)`
C. `(eV)/(h)`
D. `(h)/(eV)`

Explanation

To solve the question regarding the maximum frequency of continuous X-rays when an electron is accelerated by a voltage \( V \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: In an X-ray tube, the kinetic energy of the accelerated electrons is converted into the energy of the emitted X-ray photons. The maximum frequency of the X-rays is related to the maximum kinetic energy of the electrons.

2. **Kinetic Energy of the Electron**: The kinetic energy (KE) of an electron accelerated through a potential difference \( V \) is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron.

3. **Energy of a Photon**: The energy \( E \) of a photon is related to its frequency \( \nu \) by the equation:
\[ E = h\nu \]
where \( h \) is Planck's constant.

4. **Setting the Energies Equal**: For the maximum frequency of the emitted X-rays, we set the kinetic energy of the electron equal to the energy of the photon:
\[ eV = h\nu_{\text{max}} \]

5. **Solving for Maximum Frequency**: Rearranging the equation to solve for the maximum frequency \( \nu_{\text{max}} \):
\[ \nu_{\text{max}} = \frac{eV}{h} \]

6. **Final Expression**: Thus, the maximum frequency of continuous X-rays when an electron is accelerated by a voltage \( V \) is given by:
\[ \nu_{\text{max}} = \frac{eV}{h} \]

### Conclusion:
From the options provided, the correct answer is:
\[ \text{Option 3: } \frac{eV}{h} \]