The emission of electrons from a metal surface exposed rto light radaition of appropriate wavelength is called photoelectroic effect .The emmited electron are called photo=-weklectron work function of threshold energy may be defined as the minimum amount of energy required to ejercted electron from a most surface .According to Einstein
Maximum kinetic energy of ejected electron = Aborbed energy - Work function
`(1)/(2) mv_(max)^(2) = h(v) - h(v_(n)) = hv [(1)/(lambda) -0 (1)/(lambda_(n))]`
Where `v_(n)` and `lambda_(0)` are thereshold frequency and threshold wavelength respectively
Sopping potential : it is the miximum potential at which the photoelectric current becomes zero if `V_(0)` is the stopping potential `eV_(0) = h(v- v_(0))`
Whaich of the following is the graph between the frequency (V) of the incident radiation and the stopping potential (v) ?

Options

A.

B.

C.

D.

Answer

To solve the question regarding the relationship between the frequency of incident radiation and the stopping potential in the photoelectric effect, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**:
The stopping potential \( V_0 \) is related to the frequency \( \nu \) of the incident radiation by the equation:
\[ eV_0 = h\nu - h\nu_0 \]
where \( \nu_0 \) is the threshold frequency.

2. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ V_0 = \frac{h}{e} \nu - \frac{h}{e} \nu_0 \]
This equation is in the form of \( y = mx + c \), where:
- \( y \) is \( V_0 \)
- \( x \) is \( \nu \)
- \( m \) is the slope \( \frac{h}{e} \)
- \( c \) is the y-intercept \( -\frac{h}{e} \nu_0 \)

3. **Graph Characteristics**:
Since the equation is linear, the graph of \( V_0 \) against \( \nu \) will be a straight line. The slope of the line is positive, indicating that as the frequency increases, the stopping potential also increases.

4. **Identify the Correct Graph**:
Given that the graph is a straight line with a positive slope, we can eliminate any options that do not represent a straight line. The correct graph will show an upward trend as frequency increases.

5. **Conclusion**:
Based on the analysis, the correct option that represents the graph between the frequency of the incident radiation and the stopping potential is option C.