In photoelectric effect the slope of stop of stopping potential versus frequency of incident light for a given surface will be

Options

A. slope is different for different metals
B. threshold frequency does not vary with emitter
C. the slope is universal constant equal to (h/e)
D. stopping potential decreases with increasing frequency of incident light

Answer

To solve the problem regarding the slope of the stopping potential versus frequency of incident light in the photoelectric effect, we can follow these steps:

### Step 1: Understand the relationship between stopping potential and frequency
The stopping potential (V₀) is related to the maximum kinetic energy (KE_max) of the ejected electrons. The maximum kinetic energy can be expressed as:
\[ KE_{max} = h\nu - h\nu_0 \]
where:
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \nu_0 \) is the threshold frequency.

### Step 2: Relate kinetic energy to stopping potential
The stopping potential is defined as:
\[ V_0 = \frac{KE_{max}}{e} \]
where \( e \) is the charge of the electron. Substituting the expression for \( KE_{max} \):
\[ V_0 = \frac{h\nu - h\nu_0}{e} \]

### Step 3: Rearranging the equation
We can rearrange the equation to express \( V_0 \) in terms of \( \nu \):
\[ V_0 = \frac{h}{e} \nu - \frac{h\nu_0}{e} \]

### Step 4: Identify the slope of the graph
This equation is in the form of \( y = mx + c \), where:
- \( y \) is \( V_0 \),
- \( m \) is the slope,
- \( x \) is \( \nu \),
- \( c \) is the y-intercept.

From our equation, we can see that the slope \( m \) is:
\[ m = \frac{h}{e} \]

### Conclusion
Thus, the slope of the stopping potential versus frequency graph for a given surface is:
\[ \text{Slope} = \frac{h}{e} \]