The work function of a metal is 2.4 eV. If radiations of 4000 `A^o` on the metal, then the kinetic energy of the fastest photoelectron is:

Explanation

To solve the problem of finding the kinetic energy of the fastest photoelectron when a metal with a work function of 2.4 eV is exposed to radiation of wavelength 4000 Å, we can follow these steps:

### Step 1: Calculate the energy of the incident radiation
The energy of the radiation can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \))
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \))
- \( \lambda \) is the wavelength of the radiation in meters.

First, convert the wavelength from angstroms to meters:
\[ \lambda = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} = 4.0 \times 10^{-7} \, \text{m} \]

Now, substitute the values into the formula:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3 \times 10^8 \, \text{m/s})}{4.0 \times 10^{-7} \, \text{m}} \]
Calculating this gives:
\[ E = \frac{1.9878 \times 10^{-25}}{4.0 \times 10^{-7}} = 4.9695 \times 10^{-19} \, \text{J} \]

### Step 2: Convert the work function from eV to Joules
The work function (\( \phi \)) is given as 2.4 eV. We need to convert this to joules using the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \):
\[ \phi = 2.4 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} = 3.843 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the kinetic energy of the fastest photoelectron
The kinetic energy (\( KE \)) of the fastest photoelectron can be calculated using the equation:
\[ KE = E - \phi \]
Substituting the values we found:
\[ KE = (4.9695 \times 10^{-19} \, \text{J}) - (3.843 \times 10^{-19} \, \text{J}) = 1.1265 \times 10^{-19} \, \text{J} \]

### Step 4: Final Result
The kinetic energy of the fastest photoelectron is approximately:
\[ KE \approx 1.125 \times 10^{-19} \, \text{J} \]