X-rays of frequency u were used to irradiate sodium and copper surfaces in two separate experiments and their stopping potentials were found to be Na and V' cu If the work function of `Cu gt` work function of Na, then

Options

A. `V_(Na) gt V_(Cu)`
B. `V_(Na) -V_(Cu)`
C. `V_(Na) lt V_(Cu)`
D. `V_(Na) "and "V_(Cu)` are inversely proportional to `epsilon`

Explanation

To solve the problem, we need to analyze the relationship between the stopping potentials of sodium (Na) and copper (Cu) when irradiated with X-rays of frequency \( \nu_0 \).

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**:
The photoelectric effect states that when light (or electromagnetic radiation) hits a material, it can eject electrons from that material if the energy of the incoming photons is greater than the work function of the material.

2. **Define Work Function**:
The work function (\( \phi \)) is the minimum energy required to remove an electron from the surface of a material. For sodium, we denote it as \( \phi_{Na} \) and for copper as \( \phi_{Cu} \).

3. **Given Information**:
We know that \( \phi_{Cu} > \phi_{Na} \). This means that copper requires more energy to eject an electron compared to sodium.

4. **Photoelectric Equation**:
The maximum kinetic energy (\( K_{max} \)) of the emitted electrons can be expressed using the photoelectric equation:
\[ K_{max} = h\nu - \phi \]
where \( h \) is Planck's constant and \( \nu \) is the frequency of the incident radiation.

5. **Express Stopping Potential**:
The stopping potential (\( V \)) is related to the maximum kinetic energy of the emitted electrons. It can be expressed as:
\[ K_{max} = eV \]
where \( e \) is the charge of the electron. Thus, we can write:
\[ eV_{Na} = h\nu - \phi_{Na} \]
\[ eV_{Cu} = h\nu - \phi_{Cu} \]

6. **Relate Stopping Potentials**:
Rearranging the equations for stopping potentials gives us:
\[ V_{Na} = \frac{h\nu - \phi_{Na}}{e} \]
\[ V_{Cu} = \frac{h\nu - \phi_{Cu}}{e} \]

7. **Compare Stopping Potentials**:
Since \( \phi_{Cu} > \phi_{Na} \), we can compare the two expressions:
\[ V_{Na} = \frac{h\nu - \phi_{Na}}{e} \quad \text{and} \quad V_{Cu} = \frac{h\nu - \phi_{Cu}}{e} \]
Because \( \phi_{Cu} \) is larger than \( \phi_{Na} \), it follows that:
\[ h\nu - \phi_{Na} > h\nu - \phi_{Cu} \]
Therefore,
\[ V_{Na} > V_{Cu} \]

### Conclusion:
The stopping potential for sodium (\( V_{Na} \)) is greater than that for copper (\( V_{Cu} \)):
\[ V_{Na} > V_{Cu} \]