In a photeclectric experiment , when electromegnetic wave given by `E=E_(0)"sin"omegat` is incident , electron just ejects. When `E=E_(0)"sin"2omegat` is incident `K_("max")=K_(1)` and when E`=E_(0)"sin" 6omegat` is incident `K_("amx")=k_(2)`. When `k_(2)//k_(1)`

Answer

To solve the problem, we need to analyze the kinetic energy of electrons ejected in a photoelectric experiment under different electromagnetic wave conditions.

### Step-by-Step Solution:

1. **Understanding the Incident Wave**:
- The first electromagnetic wave is given by \( E = E_0 \sin(\omega t) \).
- The energy of the photon corresponding to this wave is \( E_1 = h \nu \), where \( \nu = \frac{\omega}{2\pi} \).
- Since the electron is just ejected, the energy of the photon equals the work function \( \phi \):
\[ E_1 = \phi \implies h \frac{\omega}{2\pi} = \phi \]

2. **Calculating Kinetic Energy for the Second Wave**:
- The second wave is \( E = E_0 \sin(2\omega t) \).
- Here, the frequency is \( \nu_1 = \frac{2\omega}{2\pi} = \frac{\omega}{\pi} \).
- The energy of the photon for this wave is:
\[ E_2 = h \nu_1 = h \frac{2\omega}{2\pi} = h \frac{\omega}{\pi} \]
- The kinetic energy \( K_1 \) of the ejected electron is given by:
\[ K_1 = E_2 - \phi = h \frac{\omega}{\pi} - h \frac{\omega}{2\pi} = h \frac{\omega}{\pi} \left(1 - \frac{1}{2}\right) = h \frac{\omega}{2\pi} \]

3. **Calculating Kinetic Energy for the Third Wave**:
- The third wave is \( E = E_0 \sin(6\omega t) \).
- The frequency is \( \nu_2 = \frac{6\omega}{2\pi} = \frac{3\omega}{\pi} \).
- The energy of the photon for this wave is:
\[ E_3 = h \nu_2 = h \frac{6\omega}{2\pi} = 3h \frac{\omega}{2\pi} \]
- The kinetic energy \( K_2 \) of the ejected electron is given by:
\[ K_2 = E_3 - \phi = 3h \frac{\omega}{2\pi} - h \frac{\omega}{2\pi} = 2h \frac{\omega}{2\pi} = h \frac{\omega}{\pi} \]

4. **Finding the Ratio of Kinetic Energies**:
- Now, we need to find the ratio \( \frac{K_2}{K_1} \):
\[ \frac{K_2}{K_1} = \frac{h \frac{\omega}{\pi}}{h \frac{\omega}{2\pi}} = \frac{h \frac{\omega}{\pi}}{h \frac{\omega}{2\pi}} = \frac{2\pi}{\pi} = 2 \]

### Final Answer:
\[ \frac{K_2}{K_1} = 2 \]