Maximum kinetic energy of photoelectrons from a metal
surface is `K_0` when wavelngth of incident light is `lambda`. If wavelength is decreased
to `lambda|2`, the maximum kinetic energy of photoelectrons becomes
(a) `=2K_0` (b)`gt2K_0` (c ) `lt2K_0`

Explanation

To solve the problem, we will use the principles of the photoelectric effect, specifically Einstein's photoelectric equation.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The maximum kinetic energy (K_max) of photoelectrons emitted from a metal surface is given by the equation:
\[ K_{max} = E_{photon} - \phi \]
where \(E_{photon}\) is the energy of the incoming photon and \(\phi\) is the work function of the metal.

2. **Photon Energy Calculation**:
The energy of a photon can be expressed in terms of its wavelength (\(\lambda\)):
\[ E_{photon} = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant and \(c\) is the speed of light.

3. **First Case (Wavelength = \(\lambda\))**:
For the initial wavelength \(\lambda\), the maximum kinetic energy \(K_0\) is given by:
\[ K_0 = \frac{hc}{\lambda} - \phi \]
Rearranging this gives:
\[ \frac{hc}{\lambda} = K_0 + \phi \tag{1} \]

4. **Second Case (Wavelength = \(\frac{\lambda}{2}\))**:
When the wavelength is decreased to \(\frac{\lambda}{2}\), the energy of the photon becomes:
\[ E'_{photon} = \frac{hc}{\frac{\lambda}{2}} = \frac{2hc}{\lambda} \]
The new maximum kinetic energy \(K'\) can be expressed as:
\[ K' = E'_{photon} - \phi = \frac{2hc}{\lambda} - \phi \]

5. **Substituting from Equation (1)**:
We can substitute \(\frac{hc}{\lambda}\) from Equation (1) into the equation for \(K'\):
\[ K' = 2\left(\frac{hc}{\lambda}\right) - \phi = 2(K_0 + \phi) - \phi \]
Simplifying this gives:
\[ K' = 2K_0 + 2\phi - \phi = 2K_0 + \phi \]

6. **Comparison of Kinetic Energies**:
Since \(\phi\) is a positive value (the work function cannot be negative), we can conclude:
\[ K' = 2K_0 + \phi > 2K_0 \]
Therefore, the maximum kinetic energy of the photoelectrons when the wavelength is decreased to \(\frac{\lambda}{2}\) is greater than \(2K_0\).

### Final Answer:
The maximum kinetic energy of photoelectrons becomes \(K' > 2K_0\).