The work function for metals `A , B` and `C` are respectively `1.92 eV , 2.0 eV` and `5 eV`. According to Einstein's equation , the metals which will emit photoelectrons for a radiation of wavelength `4100 Å` are

Options

A. None of these
B. A only
C. A and B only
D. All the three metals.

Answer

To solve the problem, we need to determine which metals (A, B, or C) will emit photoelectrons when exposed to radiation of a specific wavelength (4100 Å). We will use Einstein's photoelectric equation to find the energy of the incident radiation and compare it with the work functions of the metals.

### Step-by-Step Solution:

1. **Convert Wavelength to Energy**:
We use the formula for the energy of a photon:
\[ E = \frac{12400}{\lambda} \]
where \(E\) is the energy in electron volts (eV) and \(\lambda\) is the wavelength in angstroms (Å).

2. **Substitute the Given Wavelength**:
Given that \(\lambda = 4100 \, \text{Å}\):
\[ E = \frac{12400}{4100} \]

3. **Calculate the Energy**:
Performing the calculation:
\[ E = \frac{12400}{4100} \approx 3.02 \, \text{eV} \]

4. **Compare with Work Functions**:
The work functions for metals A, B, and C are:
- Metal A: \( \phi_A = 1.92 \, \text{eV} \)
- Metal B: \( \phi_B = 2.0 \, \text{eV} \)
- Metal C: \( \phi_C = 5 \, \text{eV} \)

We compare the energy of the incident radiation \(E \approx 3.02 \, \text{eV}\) with the work functions:
- For Metal A: \(3.02 \, \text{eV} > 1.92 \, \text{eV}\) (Photoelectrons emitted)
- For Metal B: \(3.02 \, \text{eV} > 2.0 \, \text{eV}\) (Photoelectrons emitted)
- For Metal C: \(3.02 \, \text{eV} < 5 \, \text{eV}\) (No photoelectrons emitted)

5. **Conclusion**:
Metals A and B will emit photoelectrons when exposed to the radiation of wavelength 4100 Å, while Metal C will not.

### Final Answer:
The metals which will emit photoelectrons for the radiation of wavelength 4100 Å are **A and B**.

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