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. Here are the steps to arrive at the solution:

### Step 1: Understand the Work Function
The work function (φ) of a metal is the minimum energy required to eject an electron from the surface of that metal. It is related to the threshold frequency (ν₀) by the equation:
\[ \phi = h \nu_0 \]
where \( h \) is Planck's constant.

Given that metal Y has a greater work function than metal X, we can express this as:
\[ \phi_Y > \phi_X \]
This implies:
\[ h \nu_{0Y} > h \nu_{0X} \]
Thus, the threshold frequency for metal Y is greater than that for metal X:
\[ \nu_{0Y} > \nu_{0X} \]

### Step 2: Einstein's Photoelectric Equation
The stopping potential (V) is related to the frequency of the incident light (f) by Einstein's photoelectric equation:
\[ eV = hf - \phi \]
where \( e \) is the charge of the electron.

For metals X and Y, we can write:
1. For metal X:
\[ eV_X = hf - h\nu_{0X} \]
Rearranging gives:
\[ V_X = \frac{hf}{e} - \frac{h\nu_{0X}}{e} \]

2. For metal Y:
\[ eV_Y = hf - h\nu_{0Y} \]
Rearranging gives:
\[ V_Y = \frac{hf}{e} - \frac{h\nu_{0Y}}{e} \]

### Step 3: Analyze the Graphs
From the equations derived, we can see that both \( V_X \) and \( V_Y \) are linear functions of frequency \( f \) with the same slope of \( \frac{h}{e} \). However, the y-intercepts will differ due to the different work functions (or threshold frequencies).

- The y-intercept for metal X will be at \( -\frac{h\nu_{0X}}{e} \).
- The y-intercept for metal Y will be at \( -\frac{h\nu_{0Y}}{e} \).

Since \( \nu_{0Y} > \nu_{0X} \), the y-intercept for metal Y will be lower than that for metal X.

### Step 4: Conclusion
The graph that illustrates the expected results will show:
- Both lines having the same slope (indicating the same value of \( \frac{h}{e} \)).
- The line for metal Y starting at a lower value on the V-axis (indicating a higher threshold frequency).

### Final Answer
The correct graph will depict two straight lines with the same slope, where the line for metal Y is positioned lower than that for metal X on the V-axis.

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