If the work function for a certain metal is `3.2 xx 10^(-19)` joule and it is illuminated with light of frequency `8 xx 10^(14) Hz`. The maximum kinetic energy of the photo-electrons would be
`(h = 6.63 xx 10^(-34) Js)`

Options

A. `2.1 xx 10^(-19)J`
B. `8.5 xx 10^(-9) J`
C. `5.3 xx 10^(-19)J`
D. `3.2 xx 10^(-19)J`

Answer

To find the maximum kinetic energy of the photo-electrons when a metal is illuminated with light, we can use the photoelectric effect equation:

\[ K_{max} = E_{photon} - \phi \]

where:
- \( K_{max} \) is the maximum kinetic energy of the photo-electrons,
- \( E_{photon} \) is the energy of the incoming photon,
- \( \phi \) is the work function of the metal.

### Step 1: Calculate the energy of the photon
The energy of a photon can be calculated using the formula:

\[ E_{photon} = h \nu \]

where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( \nu \) is the frequency of the light (\( 8 \times 10^{14} \, \text{Hz} \)).

Substituting the values:

\[ E_{photon} = (6.63 \times 10^{-34} \, \text{Js}) \times (8 \times 10^{14} \, \text{Hz}) \]

Calculating this gives:

\[ E_{photon} = 5.304 \times 10^{-19} \, \text{J} \]

### Step 2: Use the work function
The work function \( \phi \) is given as:

\[ \phi = 3.2 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the maximum kinetic energy
Now, we can substitute the values into the equation for maximum kinetic energy:

\[ K_{max} = E_{photon} - \phi \]

Substituting the calculated energy of the photon and the work function:

\[ K_{max} = (5.304 \times 10^{-19} \, \text{J}) - (3.2 \times 10^{-19} \, \text{J}) \]

Calculating this gives:

\[ K_{max} = 2.104 \times 10^{-19} \, \text{J} \]

### Final Answer
Thus, the maximum kinetic energy of the photo-electrons is:

\[ K_{max} \approx 2.1 \times 10^{-19} \, \text{J} \]

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