In a photoelectric experiment , the stopping potential `V_s` is plotted against the frequency v of the incident light . The resulting curve is a straight line which makes an angle `theta` with the v - axis. Then `tan theta` will be equal to (Here `E_0` = work function of the surface_

Options

A. `h/e`
B. `e/h`
C. `-phi/e`
D. `(eh)/phi`

Answer

To solve the problem, we need to analyze the relationship between the stopping potential \( V_s \) and the frequency \( \nu \) of the incident light in a photoelectric experiment.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The photoelectric effect can be described by the equation:
\[ E = \phi + eV_s \]
where:
- \( E \) is the energy of the incident photons,
- \( \phi \) is the work function of the material,
- \( e \) is the charge of the electron,
- \( V_s \) is the stopping potential.

2. **Relating Photon Energy to Frequency**:
The energy of the incident photons can also be expressed in terms of frequency:
\[ E = h\nu \]
where \( h \) is Planck's constant and \( \nu \) is the frequency of the incident light.

3. **Substituting Energy in the Equation**:
By substituting \( E \) in the photoelectric equation, we get:
\[ h\nu = \phi + eV_s \]

4. **Rearranging the Equation**:
Rearranging the equation gives:
\[ eV_s = h\nu - \phi \]
or,
\[ V_s = \frac{h}{e}\nu - \frac{\phi}{e} \]

5. **Identifying the Linear Relationship**:
The equation \( V_s = \frac{h}{e}\nu - \frac{\phi}{e} \) is in the form of a linear equation \( y = mx + c \), where:
- \( y \) is \( V_s \),
- \( x \) is \( \nu \),
- \( m \) (the slope) is \( \frac{h}{e} \),
- \( c \) (the y-intercept) is \( -\frac{\phi}{e} \).

6. **Finding the Slope**:
The slope of the line, which is \( \tan \theta \) (where \( \theta \) is the angle the line makes with the x-axis), is given by:
\[ \tan \theta = \frac{h}{e} \]

### Final Answer:
Thus, the value of \( \tan \theta \) is:
\[ \tan \theta = \frac{h}{e} \]