X-rays are used to irradiate sodium and copper surfaces in two separate experiments and stopping potential are determined. The stopping potential is

Options

A. equal in both cases
B. greater for sodium
C. greater for copper
D. infinite in both cases

Explanation

To solve the problem regarding the stopping potential when X-rays irradiate sodium and copper surfaces, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Concept**: The stopping potential (V_s) is the potential needed to stop the most energetic photoelectrons emitted from a surface when it is irradiated by light (or X-rays in this case). The stopping potential is directly related to the maximum kinetic energy (K.E.) of the emitted electrons.

2. **Energy of Irradiation**: The energy of the incident X-rays can be expressed using the equation:
\[ E = h \nu \]
where \(E\) is the energy, \(h\) is Planck's constant, and \(\nu\) is the frequency of the X-rays. Since the frequency is the same for both experiments, the energy \(E\) is the same for both sodium and copper.

3. **Kinetic Energy of Emitted Electrons**: The maximum kinetic energy of the emitted electrons can be given by the equation:
\[ K.E. = E - W \]
where \(W\) is the work function of the material.

4. **Work Functions of Sodium and Copper**: It is known that the work function of sodium (\(W_{Na}\)) is lower than that of copper (\(W_{Cu}\)):
\[ W_{Na} < W_{Cu} \]

5. **Calculating Kinetic Energies**: For sodium, the kinetic energy can be expressed as:
\[ K.E_{Na} = E - W_{Na} \]
For copper, it is:
\[ K.E_{Cu} = E - W_{Cu} \]

6. **Comparing Kinetic Energies**: Since \(W_{Na} < W_{Cu}\) and \(E\) is the same for both, we can conclude:
\[ K.E_{Na} > K.E_{Cu} \]
This means that the maximum kinetic energy of electrons emitted from sodium is greater than that from copper.

7. **Relating Kinetic Energy to Stopping Potential**: The stopping potential is related to the maximum kinetic energy by the equation:
\[ K.E = eV_s \]
where \(e\) is the charge of the electron. Therefore, the stopping potential can be expressed as:
\[ V_s = \frac{K.E}{e} \]

8. **Conclusion**: Since \(K.E_{Na} > K.E_{Cu}\), it follows that:
\[ V_{s, Na} > V_{s, Cu} \]
Thus, the stopping potential for sodium is greater than that for copper.

### Final Answer:
The stopping potential for sodium is greater than that for copper.