The work function of a metal is `1.6 xx 10^(-19) J`. When the metal surface is illuminated by the light of wavelength `6400 Å`, then the maximum kinetic energy of emitted photo - electrons will be
(Planck's constant `h = 6.4 xx 10^(-34) Js`)

Options

A. `14 xx 10^(-19) J`
B. `2.8 xx 10^(-19) J`
C. `1.4 xx 10^(-19) J`
D. `1.4 xx 10^(19) J`

Explanation

To find the maximum kinetic energy of emitted photo-electrons when a metal surface is illuminated by light of a certain wavelength, we can use the photoelectric effect equation derived from Einstein's photoelectric equation:

\[ K_{\text{max}} = E_{\text{incident}} - W \]

Where:
- \( K_{\text{max}} \) = maximum kinetic energy of the emitted photo-electrons
- \( E_{\text{incident}} \) = energy of the incident photons
- \( W \) = work function of the metal

### Step 1: Calculate the energy of the incident photons (\( E_{\text{incident}} \))

The energy of the incident photons can be calculated using the formula:

\[ E_{\text{incident}} = \frac{hc}{\lambda} \]

Where:
- \( h \) = Planck's constant = \( 6.4 \times 10^{-34} \, \text{Js} \)
- \( c \) = speed of light = \( 3 \times 10^8 \, \text{m/s} \)
- \( \lambda \) = wavelength of the light = \( 6400 \, \text{Å} = 6400 \times 10^{-10} \, \text{m} = 6.4 \times 10^{-7} \, \text{m} \)

Substituting the values:

\[ E_{\text{incident}} = \frac{(6.4 \times 10^{-34} \, \text{Js}) \times (3 \times 10^8 \, \text{m/s})}{6.4 \times 10^{-7} \, \text{m}} \]

Calculating the numerator:

\[ 6.4 \times 10^{-34} \times 3 \times 10^8 = 19.2 \times 10^{-26} \, \text{Jm} \]

Now calculating \( E_{\text{incident}} \):

\[ E_{\text{incident}} = \frac{19.2 \times 10^{-26}}{6.4 \times 10^{-7}} = 3.0 \times 10^{-19} \, \text{J} \]

### Step 2: Subtract the work function from the energy of the incident photons

Now we will use the work function \( W = 1.6 \times 10^{-19} \, \text{J} \):

\[ K_{\text{max}} = E_{\text{incident}} - W \]

Substituting the values:

\[ K_{\text{max}} = (3.0 \times 10^{-19} \, \text{J}) - (1.6 \times 10^{-19} \, \text{J}) \]

Calculating \( K_{\text{max}} \):

\[ K_{\text{max}} = 1.4 \times 10^{-19} \, \text{J} \]

### Final Answer

The maximum kinetic energy of emitted photo-electrons is:

\[ K_{\text{max}} = 1.4 \times 10^{-19} \, \text{J} \]