The work function of a surface is 3.1 eV. A photon of frequency` 1x 10^(15)`Hz. Is incident on it. Calculate the incident wavelength is photoelectric emission occur or not.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the Wavelength of the Incident Photon
We know the relationship between the speed of light (c), frequency (ν), and wavelength (λ):
\[ \lambda = \frac{c}{\nu} \]
Where:
- \( c = 3 \times 10^8 \) m/s (speed of light)
- \( \nu = 1 \times 10^{15} \) Hz (frequency of the photon)

Substituting the values:
\[ \lambda = \frac{3 \times 10^8 \text{ m/s}}{1 \times 10^{15} \text{ Hz}} \]
\[ \lambda = 3 \times 10^{-7} \text{ m} \]

### Step 2: Calculate the Energy of the Incident Photon
The energy (E) of a photon can be calculated using the formula:
\[ E = h \nu \]
Where:
- \( h = 6.63 \times 10^{-34} \) J·s (Planck's constant)
- \( \nu = 1 \times 10^{15} \) Hz (frequency of the photon)

Substituting the values:
\[ E = 6.63 \times 10^{-34} \text{ J·s} \times 1 \times 10^{15} \text{ Hz} \]
\[ E = 6.63 \times 10^{-19} \text{ J} \]

### Step 3: Convert the Energy from Joules to Electron Volts
To convert energy from Joules to electron volts (eV), we use the conversion factor:
\[ 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \]

Thus, we can convert the energy:
\[ E = \frac{6.63 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} \]
\[ E \approx 4.14 \text{ eV} \]

### Step 4: Compare the Energy of the Photon with the Work Function
The work function (Φ) of the surface is given as:
\[ Φ = 3.1 \text{ eV} \]

Now, we compare the energy of the photon (4.14 eV) with the work function (3.1 eV):
- Since \( E (4.14 \text{ eV}) > Φ (3.1 \text{ eV}) \), photoelectric emission will occur.

### Conclusion
The incident wavelength is \( 3 \times 10^{-7} \) m, and since the energy of the photon exceeds the work function, photoelectric emission will occur.

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