The work function of Silver and sodium are `4.6 and 2.3 eV, ` respectively . The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is

Explanation

To solve the problem, we need to find the ratio of the slopes of the stopping potential versus frequency plot for silver and sodium. The work functions for silver and sodium are given as 4.6 eV and 2.3 eV, respectively.

### Step-by-Step Solution:

1. **Understand the Photoelectric Equation**:
The photoelectric effect can be described by the equation:
\[ E_k = h\nu - \phi \]
where \(E_k\) is the kinetic energy of the emitted electrons, \(h\) is Planck's constant, \(\nu\) is the frequency of the incident light, and \(\phi\) is the work function of the material.

2. **Relate Kinetic Energy to Stopping Potential**:
The stopping potential \(V_0\) is related to the kinetic energy by:
\[ E_k = eV_0 \]
where \(e\) is the charge of the electron. Thus, we can rewrite the photoelectric equation as:
\[ eV_0 = h\nu - \phi \]

3. **Rearranging the Equation**:
Rearranging gives us:
\[ V_0 = \frac{h}{e}\nu - \frac{\phi}{e} \]
This equation is in the form of \(y = mx + c\), where:
- \(y\) is \(V_0\)
- \(x\) is \(\nu\)
- \(m\) (the slope) is \(\frac{h}{e}\)
- \(c\) (the y-intercept) is \(-\frac{\phi}{e}\)

4. **Finding the Slope**:
From the equation, the slope \(m\) is constant and equal to \(\frac{h}{e}\). This means that the slope does not depend on the material but rather on fundamental constants.

5. **Calculating the Ratio of Slopes**:
Since the slope for both silver and sodium is \(\frac{h}{e}\), we can write:
\[ \text{Slope for Silver} = \frac{h}{e} \]
\[ \text{Slope for Sodium} = \frac{h}{e} \]
Therefore, the ratio of the slopes is:
\[ \text{Ratio} = \frac{\text{Slope for Silver}}{\text{Slope for Sodium}} = \frac{\frac{h}{e}}{\frac{h}{e}} = 1 \]

### Final Answer:
The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is **1**.