Find the maximum velocity of photoelectrons emitted by radiation of frequency `3 xx 10^(15)` Hz from a photoelectric surface having a work function of 4.0 eV. Given `h=6.6 xx 10^(-34)` Js and 1 eV = `1.6 xx 10^(-19)` J.

Answer

To find the maximum velocity of photoelectrons emitted from a photoelectric surface, we will use Einstein's photoelectric equation and the kinetic energy formula.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Frequency of radiation, \( \nu = 3 \times 10^{15} \, \text{Hz} \)
- Work function, \( \phi = 4.0 \, \text{eV} \)
- Planck's constant, \( h = 6.6 \times 10^{-34} \, \text{Js} \)
- Conversion factor, \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \)
- Mass of electron, \( m = 9.1 \times 10^{-31} \, \text{kg} \)

2. **Calculate the Energy of the Incident Photons**:
The energy of the incident photons can be calculated using the formula:
\[ E = h \nu \]
Substituting the values:
\[ E = (6.6 \times 10^{-34} \, \text{Js}) \times (3 \times 10^{15} \, \text{Hz}) = 1.98 \times 10^{-18} \, \text{J} \]

3. **Convert Work Function from eV to Joules**:
The work function in Joules is calculated as:
\[ \phi = 4.0 \, \text{eV} \times (1.6 \times 10^{-19} \, \text{J/eV}) = 6.4 \times 10^{-19} \, \text{J} \]

4. **Calculate the Kinetic Energy of the Emitted Electrons**:
The kinetic energy (KE) of the emitted electrons is given by:
\[ KE = E - \phi \]
Substituting the values:
\[ KE = (1.98 \times 10^{-18} \, \text{J}) - (6.4 \times 10^{-19} \, \text{J}) = 1.34 \times 10^{-18} \, \text{J} \]

5. **Use the Kinetic Energy to Find Maximum Velocity**:
The kinetic energy is also given by:
\[ KE = \frac{1}{2} mv^2 \]
Rearranging for \( v \):
\[ v = \sqrt{\frac{2 \times KE}{m}} \]
Substituting the values:
\[ v = \sqrt{\frac{2 \times (1.34 \times 10^{-18} \, \text{J})}{9.1 \times 10^{-31} \, \text{kg}}} \]

6. **Calculate the Maximum Velocity**:
\[ v = \sqrt{\frac{2.68 \times 10^{-18}}{9.1 \times 10^{-31}}} = \sqrt{2.94 \times 10^{12}} \approx 1.71 \times 10^6 \, \text{m/s} \]

### Final Answer:
The maximum velocity of the photoelectrons emitted is approximately \( 1.71 \times 10^6 \, \text{m/s} \).

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