When photons of energy `(hc)/(lamda)` fall on a metal surface, photoelectrons are ejected from it. If the work function of the surfacei s `hupsilon_0`, then

Options

A. maximum kinetic energy of the electron is `[((hc)/(lamda))-hv_0]`
B. maximum kinetic energy of the photoelectron is equal to `((hc)/(lamda))`
C. minimum KE of the photoelectron is zero
D. minimum kinetic energy of the photoelectron is equal to `(hc)/(lamda)`

Explanation

To solve the problem, we will use the concept of the photoelectric effect, which is described by Einstein's photoelectric equation. The equation relates the energy of the incident photons to the work function of the metal and the maximum kinetic energy of the emitted photoelectrons.

### Step-by-Step Solution:

1. **Identify the Energy of the Incident Photons**:
The energy of the incident photons can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the incident light.

2. **Write Einstein's Photoelectric Equation**:
According to Einstein's photoelectric equation:
\[ E = W + KE_{\text{max}} \]
where \( W \) is the work function of the metal and \( KE_{\text{max}} \) is the maximum kinetic energy of the emitted photoelectrons.

3. **Substitute the Work Function**:
The work function \( W \) is given as \( h\nu_0 \), where \( \nu_0 \) is the threshold frequency of the metal. Thus, we can rewrite the equation as:
\[ \frac{hc}{\lambda} = h\nu_0 + KE_{\text{max}} \]

4. **Rearrange to Find Maximum Kinetic Energy**:
To find the maximum kinetic energy of the photoelectrons, rearrange the equation:
\[ KE_{\text{max}} = \frac{hc}{\lambda} - h\nu_0 \]

5. **Interpret the Results**:
- The maximum kinetic energy of the photoelectrons is given by:
\[ KE_{\text{max}} = \frac{hc}{\lambda} - h\nu_0 \]
- The minimum kinetic energy of the photoelectrons occurs when the energy of the incident photons is just equal to the work function, leading to:
\[ KE_{\text{min}} = 0 \]
This happens when \( \frac{hc}{\lambda} = h\nu_0 \).

### Final Answer:
- The maximum kinetic energy of the emitted photoelectrons is:
\[ KE_{\text{max}} = \frac{hc}{\lambda} - h\nu_0 \]
- The minimum kinetic energy of the photoelectrons is:
\[ KE_{\text{min}} = 0 \]