The photoelectric work function of a surface is 2.2 eV. The maximum kinetic energy of photoelectrons emitted when light of wavelength 6200 `Å` is incident on the surface, is

Options

A. 0.4 eV
B. 1.2 eV
C. 1.6 eV
D. Photoelectrons are not emitted.

Answer

To solve the problem of finding the maximum kinetic energy of photoelectrons emitted when light of wavelength 6200 Å is incident on a surface with a work function of 2.2 eV, we can follow these steps:

### Step 1: Convert Wavelength to Meters
First, we need to convert the wavelength from angstroms to meters for our calculations.
1 Å = \(10^{-10}\) m, so:
\[ \text{Wavelength} (\lambda) = 6200 \, \text{Å} = 6200 \times 10^{-10} \, \text{m} = 6.2 \times 10^{-7} \, \text{m} \]

### Step 2: Calculate the Energy of the Incident Photon
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) (Planck's constant) = \(4.1357 \times 10^{-15} \, \text{eV s}\)
- \(c\) (speed of light) = \(3 \times 10^8 \, \text{m/s}\)

Substituting the values into the formula:
\[ E = \frac{(4.1357 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{6.2 \times 10^{-7} \, \text{m}} \]
Calculating this gives:
\[ E = \frac{1.240 \times 10^{-6} \, \text{eV m}}{6.2 \times 10^{-7} \, \text{m}} \approx 2.0 \, \text{eV} \]

### Step 3: Calculate the Maximum Kinetic Energy of the Photoelectrons
The maximum kinetic energy (K.E.) of the emitted photoelectrons can be calculated using the equation:
\[ K.E. = E - \phi \]
where \(\phi\) is the work function of the material. Given that \(\phi = 2.2 \, \text{eV}\):
\[ K.E. = 2.0 \, \text{eV} - 2.2 \, \text{eV} = -0.2 \, \text{eV} \]

### Step 4: Analyze the Result
Since the kinetic energy is negative, this indicates that the energy of the incident photons is not sufficient to overcome the work function of the surface. Therefore, no photoelectrons will be emitted.

### Final Answer
The maximum kinetic energy of the photoelectrons emitted is \(0 \, \text{eV}\) (no photoelectrons are emitted).

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