Light of `3,000Å` is incident on a surface of work function 4 eV . Calculate the potential difference required to stop the photoelectric emission.

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Incident Energy
The energy of the incident light can be calculated using the formula:
\[ E = \frac{12400 \, \text{eV} \cdot \text{Å}}{\lambda} \]
where \( \lambda \) is the wavelength in Angstroms.

Given:
\[ \lambda = 3000 \, \text{Å} \]

Substituting the value of \( \lambda \):
\[ E = \frac{12400 \, \text{eV} \cdot \text{Å}}{3000 \, \text{Å}} = \frac{12400}{3000} \, \text{eV} \approx 4.13 \, \text{eV} \]

### Step 2: Determine the Maximum Kinetic Energy
The maximum kinetic energy (K.E.) of the emitted electrons can be found using the photoelectric equation:
\[ \text{K.E.} = E - \phi \]
where \( \phi \) is the work function of the material.

Given:
\[ \phi = 4 \, \text{eV} \]

Substituting the values:
\[ \text{K.E.} = 4.13 \, \text{eV} - 4 \, \text{eV} = 0.13 \, \text{eV} \]

### Step 3: Calculate the Stopping Potential
The stopping potential (\( V_s \)) can be calculated using the relationship:
\[ \text{K.E.} = e \cdot V_s \]
where \( e \) is the charge of an electron (1 eV corresponds to 1 volt for an electron).

Thus,
\[ V_s = \text{K.E.} = 0.13 \, \text{V} \]

### Final Answer
The potential difference required to stop the photoelectric emission is:
\[ V_s = 0.13 \, \text{V} \]

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