A light whose frequency is equal to `6xx10^(14)Hz` is incident on a metal whose work function is `2eV(h=6.63xx10^(-34)Js,1eV=1.6xx10^(-19)J)`. The maximum energy of electrons emitted will be:

Options

A. 2.49 eV
B. 4.49 eV
C. 0.49 eV
D. 5.49 eV

Answer

To solve the problem, we need to find the maximum energy of the electrons emitted when light of a certain frequency is incident on a metal with a given work function. We will use the photoelectric effect equation:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Frequency of light, \( \nu = 6 \times 10^{14} \, \text{Hz} \)
- Work function, \( \phi = 2 \, \text{eV} \)
- Planck's constant, \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Conversion factor, \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \)

2. **Calculate the Energy of the Incident Light**:
The energy \( E \) of the light can be calculated using the formula:
\[ E = h \nu \]
Substituting the values:
\[ E = (6.63 \times 10^{-34} \, \text{Js}) \times (6 \times 10^{14} \, \text{Hz}) \]

3. **Perform the Calculation**:
\[ E = 3.978 \times 10^{-19} \, \text{J} \]

4. **Convert Energy from Joules to Electron Volts**:
To convert the energy from Joules to electron volts, we use the conversion factor:
\[ E (\text{in eV}) = \frac{E (\text{in J})}{1.6 \times 10^{-19} \, \text{J/eV}} \]
Substituting the value:
\[ E (\text{in eV}) = \frac{3.978 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.486 \, \text{eV} \]

5. **Calculate the Maximum Energy of the Emitted Electrons**:
The maximum energy \( E_{\text{max}} \) of the emitted electrons is given by:
\[ E_{\text{max}} = E - \phi \]
Substituting the values:
\[ E_{\text{max}} = 2.486 \, \text{eV} - 2 \, \text{eV} = 0.486 \, \text{eV} \]

6. **Final Answer**:
The maximum energy of the emitted electrons is approximately:
\[ E_{\text{max}} \approx 0.49 \, \text{eV} \]

### Final Result:
The maximum energy of electrons emitted will be **0.49 eV**.