The work function of a metal is `2.5eV`. The maximum kinetic energy of the photoelectrons emitted if a radiof wavelength `3000A^(0)` falls on it is `(h=6.63xx10^(-34)Js & c=3xx10^(8)m//s)`

Options

A. `1.12xx10^(-19) J`
B. `4.8xx10^(-19) J`
C. `3.2xx10^(-19) J`
D. `2.61xx10^(-19) J`

Explanation

To solve the problem, we will use the photoelectric effect equation derived from Einstein's theory, which states that the maximum kinetic energy (K.E.) of the emitted photoelectrons is given by:

\[ K.E. = E_{incident} - \phi \]

where:
- \( E_{incident} \) is the energy of the incident photon,
- \( \phi \) is the work function of the metal.

### Step 1: Convert the work function from eV to Joules
The work function \( \phi \) is given as \( 2.5 \, eV \). We need to convert this to Joules using the conversion factor \( 1 \, eV = 1.6 \times 10^{-19} \, J \).

\[ \phi = 2.5 \, eV \times 1.6 \times 10^{-19} \, J/eV = 4.0 \times 10^{-19} \, J \]

### Step 2: Calculate the energy of the incident photon
The energy of the incident photon can be calculated using the formula:

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

where:
- \( h = 6.63 \times 10^{-34} \, J \cdot s \) (Planck's constant),
- \( c = 3 \times 10^{8} \, m/s \) (speed of light),
- \( \lambda = 3000 \, \text{Å} = 3000 \times 10^{-10} \, m = 3.0 \times 10^{-7} \, m \).

Substituting the values:

\[ E_{incident} = \frac{(6.63 \times 10^{-34} \, J \cdot s)(3 \times 10^{8} \, m/s)}{3.0 \times 10^{-7} \, m} \]

Calculating the numerator:

\[ E_{incident} = \frac{1.989 \times 10^{-25} \, J \cdot m}{3.0 \times 10^{-7} \, m} = 6.63 \times 10^{-19} \, J \]

### Step 3: Calculate the maximum kinetic energy of the emitted photoelectrons
Now, we can find the maximum kinetic energy using the values we calculated:

\[ K.E. = E_{incident} - \phi \]

Substituting the values:

\[ K.E. = (6.63 \times 10^{-19} \, J) - (4.0 \times 10^{-19} \, J) = 2.63 \times 10^{-19} \, J \]

### Step 4: Convert the kinetic energy back to eV
To convert the kinetic energy back to eV:

\[ K.E. = \frac{2.63 \times 10^{-19} \, J}{1.6 \times 10^{-19} \, J/eV} \approx 1.64 \, eV \]

### Conclusion
The maximum kinetic energy of the photoelectrons emitted is approximately \( 1.64 \, eV \).