The work function of a metal is `4.0 eV`. If the metal is irradiated with radiation of wavelength `200` nm, then the maximum kinetic energy of the photoelectrons would be about.

Options

A. `6.4 xx 10^-19 J`
B. `3.5 xx 10^-19 J`
C. `1.0 xx 10^-18 J`
D. `2.0 xx 10^-19 J`

Explanation

To find the maximum kinetic energy of photoelectrons emitted from a metal when irradiated with light, we can use the photoelectric equation:

\[ KE_{\text{max}} = E_{\text{photon}} - \phi \]

Where:
- \( KE_{\text{max}} \) is the maximum kinetic energy of the emitted photoelectrons.
- \( E_{\text{photon}} \) is the energy of the incident photon.
- \( \phi \) is the work function of the metal.

### Step 1: Calculate the energy of the incident photon

The energy of the photon can be calculated using the equation:

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

Where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)).
- \( c \) is the speed of light (\( 3.0 \times 10^8 \, \text{m/s} \)).
- \( \lambda \) is the wavelength of the incident radiation (in meters).

Given:
- \( \lambda = 200 \, \text{nm} = 200 \times 10^{-9} \, \text{m} \)

Now substituting the values:

\[ E_{\text{photon}} = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.0 \times 10^8 \, \text{m/s})}{200 \times 10^{-9} \, \text{m}} \]

Calculating this gives:

\[ E_{\text{photon}} = \frac{1.9878 \times 10^{-25}}{200 \times 10^{-9}} \]
\[ E_{\text{photon}} = 9.939 \times 10^{-19} \, \text{J} \]

### Step 2: Convert the work function from eV to Joules

The work function \( \phi \) is given as \( 4.0 \, \text{eV} \). To convert this to Joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

\[ \phi = 4.0 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 6.4 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the maximum kinetic energy of the photoelectrons

Now we can find the maximum kinetic energy using the photoelectric equation:

\[ KE_{\text{max}} = E_{\text{photon}} - \phi \]

Substituting the values we calculated:

\[ KE_{\text{max}} = (9.939 \times 10^{-19} \, \text{J}) - (6.4 \times 10^{-19} \, \text{J}) \]
\[ KE_{\text{max}} = 3.539 \times 10^{-19} \, \text{J} \]

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

To convert the kinetic energy back to eV:

\[ KE_{\text{max}} = \frac{3.539 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \approx 2.21 \, \text{eV} \]

### Final Answer

The maximum kinetic energy of the photoelectrons would be approximately \( 2.21 \, \text{eV} \).

---