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. `2.1xx10^(-19)J`
B. `3.2xx10^(-19)J`
C. `5.3xx10^(-19)J`
D. `8.5xx10^(-19)J`

Answer

To find the maximum kinetic energy of the photoelectrons, we can use the photoelectric equation:

\[ K.E. = h\nu - \phi \]

where:
- \( K.E. \) is the maximum kinetic energy of the emitted photoelectrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \phi \) (or \( W \)) is the work function of the metal.

### Step 1: Identify the given values
- Work function \( \phi = 3.2 \times 10^{-19} \) joules
- Frequency \( \nu = 8 \times 10^{14} \) Hz
- Planck's constant \( h = 6.63 \times 10^{-34} \) Js

### Step 2: Calculate the energy of the incident photons
Using the formula \( E = h\nu \):

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

Calculating this gives:

\[ E = 6.63 \times 8 \times 10^{-20} = 53.04 \times 10^{-20} = 5.304 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the maximum kinetic energy
Now, substituting the values into the photoelectric equation:

\[ K.E. = E - \phi = (5.304 \times 10^{-19} \, \text{J}) - (3.2 \times 10^{-19} \, \text{J}) \]

Calculating this gives:

\[ K.E. = (5.304 - 3.2) \times 10^{-19} = 2.104 \times 10^{-19} \, \text{J} \]

### Step 4: Final answer
Thus, the maximum kinetic energy of the photoelectrons is:

\[ K.E. \approx 2.1 \times 10^{-19} \, \text{J} \]