In a photoelectric emission, electrons are ejected from metals X and Y by light of frequency f. The potential difference V required to stop the electrons is measured for various frequencies.. If Y has a greater work function than X, which graph illustrates the expected results?

Options

A.
B.
C.
D.

Explanation

To solve the problem regarding the photoelectric emission from metals X and Y, we need to analyze the relationship between the stopping potential (V) and the frequency (f) of the incident light, taking into account the work functions of the two metals.

### Step-by-Step Solution:

1. **Understand the Photoelectric Effect**: The photoelectric effect states that when light of sufficient frequency strikes a metal surface, it can eject electrons from that surface. The kinetic energy of the emitted electrons is given by the equation:
\[ KE = hf - \phi \]
where \( KE \) is the kinetic energy of the emitted electrons, \( hf \) is the energy of the incident photons, and \( \phi \) is the work function of the metal.

2. **Relate Stopping Potential to Frequency**: The stopping potential \( V \) is related to the kinetic energy of the emitted electrons. The stopping potential can be expressed as:
\[ eV = KE = hf - \phi \]
Rearranging gives:
\[ V = \frac{hf}{e} - \frac{\phi}{e} \]
This shows that the stopping potential \( V \) is linearly related to the frequency \( f \) with a slope of \( \frac{h}{e} \) and a y-intercept of \( -\frac{\phi}{e} \).

3. **Identify Work Functions**: Given that metal Y has a greater work function than metal X, we denote:
- Work function of metal X: \( \phi_X \)
- Work function of metal Y: \( \phi_Y \) where \( \phi_Y > \phi_X \)

4. **Write the Stopping Potential Equations**:
- For metal X:
\[ V_X = \frac{hf}{e} - \frac{\phi_X}{e} \]
- For metal Y:
\[ V_Y = \frac{hf}{e} - \frac{\phi_Y}{e} \]

5. **Analyze the Graphs**: Both equations have the same slope \( \frac{h}{e} \), indicating that the lines will be parallel. However, since \( \phi_Y > \phi_X \), the y-intercept of the line for metal Y will be lower (more negative) than that for metal X. This means that the line for metal Y will start lower on the y-axis compared to the line for metal X.

6. **Determine the Correct Graph**: The graph that illustrates these relationships will show two parallel lines where the line corresponding to metal Y is below the line corresponding to metal X. The correct graph will depict this situation accurately.

### Conclusion:
Based on the analysis, the graph that illustrates the expected results is option A, where the line for metal Y (with the greater work function) is lower than that for metal X.