When a high frequency electromagnetic radiation is incident on a metallic surface, electrons are emitted from the surface. Energy of emitted photoelectrons depends only on the frequency of incident electromagnetic radiation and the number of emitted electrons depends only on the intensity of incident light.
Einstein's protoelectron equation `[K_(max)=hv-phi}` correctly explains the PE, where `upsilon=` frequency of incident light and `phi=` work function.
Q. The slope of the graph between stopping potential and frequency is [here h is the Planck's constant and e is the charge of an electron] is

Options

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

Explanation

To solve the problem, we need to find the slope of the graph between stopping potential (V₀) and frequency (ν) based on Einstein's photoelectric equation. Let's break it down step by step.

### Step 1: Understand Einstein's Photoelectric Equation
Einstein's photoelectric equation is given by:
\[ K_{max} = h\nu - \phi \]
where:
- \( K_{max} \) is the maximum kinetic energy of the emitted photoelectrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \phi \) is the work function of the metal.

### Step 2: Relate Kinetic Energy to Stopping Potential
The maximum kinetic energy of the emitted photoelectrons can also be expressed in terms of stopping potential (V₀):
\[ K_{max} = eV₀ \]
where \( e \) is the charge of an electron.

### Step 3: Set the Two Equations Equal
From the two expressions for \( K_{max} \), we can equate them:
\[ eV₀ = h\nu - \phi \]

### Step 4: Rearrange the Equation
Rearranging the equation to express stopping potential (V₀) in terms of frequency (ν):
\[ V₀ = \frac{h}{e}\nu - \frac{\phi}{e} \]

### Step 5: Identify the Slope
The equation \( V₀ = \frac{h}{e}\nu - \frac{\phi}{e} \) is in the form of a straight line \( y = mx + c \), where:
- \( y \) corresponds to \( V₀ \),
- \( x \) corresponds to \( \nu \),
- \( m \) (the slope) is \( \frac{h}{e} \),
- \( c \) (the y-intercept) is \( -\frac{\phi}{e} \).

### Conclusion
Thus, the slope of the graph between stopping potential and frequency is:
\[ \text{Slope} = \frac{h}{e} \]

### Final Answer
The slope of the graph between stopping potential and frequency is \( \frac{h}{e} \).

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