It takes 4.6 eV remove one of the least tightly bound electrons from a metal surface . When monochromatic photons strike energy from zero to 2.2 eV are ejected . What is the energy of the incident photons ?

Options

A. 2.4 eV
B. 2.2 eV
C. 6.8 eV
D. 4.6 eV

Answer

To solve the problem, we will use the concept of the photoelectric effect, which is described by Einstein's photoelectric equation:

\[ KE_{max} = h\nu - \phi \]

Where:
- \( KE_{max} \) is the maximum kinetic energy of the ejected electrons.
- \( h\nu \) is the energy of the incident photons.
- \( \phi \) is the work function of the metal.

### Step-by-Step Solution:

1. **Identify the Work Function (\( \phi \))**:
The problem states that it takes 4.6 eV to remove one of the least tightly bound electrons from the metal surface. Therefore, the work function \( \phi \) is:
\[ \phi = 4.6 \, \text{eV} \]

2. **Identify the Maximum Kinetic Energy (\( KE_{max} \))**:
The problem mentions that the maximum kinetic energy of the ejected electrons is 2.2 eV. Thus:
\[ KE_{max} = 2.2 \, \text{eV} \]

3. **Apply the Photoelectric Equation**:
We can substitute the values of \( KE_{max} \) and \( \phi \) into the photoelectric equation:
\[ KE_{max} = h\nu - \phi \]
Rearranging this gives us:
\[ h\nu = KE_{max} + \phi \]

4. **Substitute the Known Values**:
Now we substitute the known values into the equation:
\[ h\nu = 2.2 \, \text{eV} + 4.6 \, \text{eV} \]

5. **Calculate the Energy of the Incident Photons**:
Performing the addition:
\[ h\nu = 6.8 \, \text{eV} \]

6. **Conclusion**:
The energy of the incident photons is:
\[ h\nu = 6.8 \, \text{eV} \]

### Final Answer:
The energy of the incident photons is **6.8 eV**.