The K.E. of an electron emitted from tungsten surface is `3.06 eV`. What voltage would be required to bring the electron to set.

Explanation

To solve the problem of finding the voltage required to bring an electron emitted from a tungsten surface to rest, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data**:
- The kinetic energy (K.E.) of the emitted electron is given as \(3.06 \, \text{eV}\).

2. **Convert Kinetic Energy to Joules**:
- The conversion factor from electron volts to joules is \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\).
- Therefore, we convert the kinetic energy:
\[ K.E. = 3.06 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 4.896 \times 10^{-19} \, \text{J} \]

3. **Relate Kinetic Energy to Voltage**:
- The voltage \(V\) required to bring the electron to rest can be calculated using the formula:
\[ V = \frac{K.E.}{e} \]
- Here, \(e\) is the charge of the electron, which is approximately \(1.6 \times 10^{-19} \, \text{C}\).

4. **Calculate the Voltage**:
- Substituting the values into the equation:
\[ V = \frac{4.896 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{C}} = 3.06 \, \text{V} \]

5. **Final Answer**:
- The voltage required to bring the electron to rest is \(3.06 \, \text{V}\).

### Summary of the Solution:
The voltage required to bring the emitted electron from tungsten to rest is \(3.06 \, \text{V}\).