The work function of aluminium is `4.2 eV`. If two photons , each of energy `3.5 eV` strike an electron of aluminium , then emission of electrons will be

Options

A. Possible
B. Not possible
C. Data is incomplete
D. Depend upon the density of the surface

Explanation

To determine whether electrons will be emitted from aluminium when two photons, each with an energy of 3.5 eV, strike it, we need to analyze the situation step by step.

### Step-by-Step Solution:

1. **Identify the Work Function**:
The work function (Φ) of aluminium is given as 4.2 eV. The work function is the minimum energy required to remove an electron from the surface of the metal.

2. **Calculate the Total Energy of the Photons**:
Each photon has an energy of 3.5 eV. Since two photons are striking the electron, we can calculate the total energy provided by the two photons:
\[ \text{Total Energy} = \text{Energy of Photon 1} + \text{Energy of Photon 2} = 3.5 \, \text{eV} + 3.5 \, \text{eV} = 7.0 \, \text{eV} \]

3. **Compare Total Energy with Work Function**:
Now, we compare the total energy of the two photons with the work function of aluminium:
\[ \text{Total Energy} = 7.0 \, \text{eV} \quad \text{and} \quad \text{Work Function} = 4.2 \, \text{eV} \]
Since 7.0 eV (total energy) is greater than 4.2 eV (work function), it indicates that the energy provided by the photons is sufficient to overcome the work function.

4. **Conclusion on Electron Emission**:
Since the total energy of the two photons exceeds the work function, electrons will be emitted from the aluminium surface. The excess energy (7.0 eV - 4.2 eV = 2.8 eV) will be converted into kinetic energy of the emitted electrons.

### Final Answer:
Yes, electrons will be emitted from aluminium when two photons, each of energy 3.5 eV, strike it.

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