A non-monochromatic light is used in an experiment on photoelectric effect. The stopping potential

Options

A. is related to the mean wavelength
B. is related to the longest wavelength
C. is related to the shortest wavelength
D. is not related to the wavelength

Answer

To solve the question regarding the stopping potential in an experiment involving non-monochromatic light and the photoelectric effect, we will follow these steps:

### Step 1: Understand the Photoelectric Effect
The photoelectric effect is the phenomenon where electrons are emitted from a material (usually a metal) when it is exposed to light of sufficient energy. The energy of the incident light is crucial in determining whether electrons will be emitted.

### Step 2: Use Einstein's Photoelectric Equation
Einstein's photoelectric equation is given by:
\[ E = \phi_0 + K_{\text{max}} \]
Where:
- \( E \) is the energy of the incident light (in joules),
- \( \phi_0 \) is the work function of the material (the minimum energy required to remove an electron),
- \( K_{\text{max}} \) is the maximum kinetic energy of the emitted electrons.

### Step 3: Relate Energy to Stopping Potential
The maximum kinetic energy of the emitted electrons can also be expressed in terms of stopping potential \( V_0 \):
\[ K_{\text{max}} = eV_0 \]
Where \( e \) is the charge of the electron.

### Step 4: Substitute the Energy Expression
The energy of the incident light can be expressed in terms of its wavelength \( \lambda \):
\[ E = h\nu = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \nu \) is the frequency of the light.

### Step 5: Combine the Equations
Substituting the expressions for energy into Einstein's equation gives:
\[ \frac{hc}{\lambda} = \phi_0 + eV_0 \]
From this, we can rearrange the equation to find the stopping potential:
\[ eV_0 = \frac{hc}{\lambda} - \phi_0 \]

### Step 6: Analyze the Relationship
From the equation \( eV_0 = \frac{hc}{\lambda} - \phi_0 \), we can see that:
- The stopping potential \( V_0 \) is directly related to the wavelength \( \lambda \).
- As the wavelength \( \lambda \) decreases (which means the energy of the light increases), the stopping potential \( V_0 \) increases.

### Step 7: Conclusion
Since the stopping potential is associated with the maximum energy of the incident light, and maximum energy corresponds to the minimum wavelength, we conclude that:
- The stopping potential \( V_0 \) is related to the shortest wavelength of the incident light.

### Final Answer
Thus, the correct option is that the stopping potential is related to the shortest wavelength of the incident light.

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