Representing the stopping potential V along X - axis and `1 / lamda` along Y - axis for a given photocathode , slope of curve for photoelectric effect is

Options

A. `(nc)/e`
B. `(he)/c`
C. `e/(hc)`
D. `c/("ne")`

Explanation

To find the slope of the curve representing the relationship between the stopping potential (V) and \( \frac{1}{\lambda} \) for the photoelectric effect, we can start from the photoelectric equation. Let's break it down step by step.

### Step-by-Step Solution:

1. **Understand the Photoelectric Equation**:
The maximum kinetic energy (KE) of the emitted electrons can be expressed as:
\[ KE = eV_s = \frac{hc}{\lambda} - \phi \]
where:
- \( e \) is the charge of the electron,
- \( V_s \) is the stopping potential,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light,
- \( \phi \) is the work function of the material.

2. **Rearranging the Equation**:
Rearranging the equation gives:
\[ eV_s = \frac{hc}{\lambda} - \phi \]
This can be rewritten as:
\[ \frac{hc}{\lambda} = eV_s + \phi \]

3. **Expressing \( \frac{1}{\lambda} \)**:
Dividing the entire equation by \( hc \) gives:
\[ \frac{1}{\lambda} = \frac{eV_s}{hc} + \frac{\phi}{hc} \]

4. **Identifying the Linear Relationship**:
This equation can be expressed in the form of \( y = mx + b \), where:
- \( y = \frac{1}{\lambda} \)
- \( x = V_s \)
- \( m = \frac{e}{hc} \) (the slope)
- \( b = \frac{\phi}{hc} \) (the y-intercept)

5. **Conclusion**:
The slope of the curve when plotting \( V_s \) against \( \frac{1}{\lambda} \) is:
\[ \text{slope} = \frac{e}{hc} \]

### Final Answer:
The slope of the curve for the photoelectric effect, when stopping potential \( V \) is plotted along the x-axis and \( \frac{1}{\lambda} \) is plotted along the y-axis, is \( \frac{e}{hc} \).

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